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x^(2)+8x+6=0 What are the solutions to the given equation? Choose 1 answer: (A) {:[x=-8-sqrt10" and "],[x=-8+sqrt10]:} (B) {:[x=-8-sqrt7" and "],[x=-8+sqrt7]:} (c) {:[x=-4-sqrt10" and "],[x=-4+sqrt10]:} (D) {:[x=-4-sqrt7" and "],[x=-4+sqrt7]:}

x2+8x+6=0 x^{2}+8 x+6=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A)\newlinex=−8−10 and x=−8+10 \begin{array}{l} x=-8-\sqrt{10} \text { and } \\ x=-8+\sqrt{10} \end{array} \newline(B)\newlinex=−8−7 and x=−8+7 \begin{array}{l} x=-8-\sqrt{7} \text { and } \\ x=-8+\sqrt{7} \end{array} \newline(C)\newlinex=−4−10 and x=−4+10 \begin{array}{l} x=-4-\sqrt{10} \text { and } \\ x=-4+\sqrt{10} \end{array} \newline(D)\newlinex=−4−7 and x=−4+7 \begin{array}{l} x=-4-\sqrt{7} \text { and } \\ x=-4+\sqrt{7} \end{array}

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Q. x2+8x+6=0 x^{2}+8 x+6=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A)\newlinex=−8−10 and x=−8+10 \begin{array}{l} x=-8-\sqrt{10} \text { and } \\ x=-8+\sqrt{10} \end{array} \newline(B)\newlinex=−8−7 and x=−8+7 \begin{array}{l} x=-8-\sqrt{7} \text { and } \\ x=-8+\sqrt{7} \end{array} \newline(C)\newlinex=−4−10 and x=−4+10 \begin{array}{l} x=-4-\sqrt{10} \text { and } \\ x=-4+\sqrt{10} \end{array} \newline(D)\newlinex=−4−7 and x=−4+7 \begin{array}{l} x=-4-\sqrt{7} \text { and } \\ x=-4+\sqrt{7} \end{array}
  1. Identify equation type: Identify the type of equation we are dealing with.\newlineThe given equation is a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0.
  2. Apply quadratic formula: Apply the quadratic formula to find the solutions for xx.\newlineThe quadratic formula is x=−b±b2−4ac2ax = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}, where a=1a = 1, b=8b = 8, and c=6c = 6.
  3. Calculate discriminant: Calculate the discriminant b2−4acb^2 - 4ac.\newlineDiscriminant = (8)2−4(1)(6)(8)^2 - 4(1)(6) = 64−2464 - 24 = 4040.
  4. Calculate solutions: Calculate the two solutions using the quadratic formula.\newlineFirst solution: x=−8−402⋅1=−8−402x = \frac{{-8 - \sqrt{40}}}{{2 \cdot 1}} = \frac{{-8 - \sqrt{40}}}{2}.\newlineSecond solution: x=−8+402⋅1=−8+402x = \frac{{-8 + \sqrt{40}}}{{2 \cdot 1}} = \frac{{-8 + \sqrt{40}}}{2}.
  5. Simplify square root: Simplify the square root of 4040.40\sqrt{40} can be simplified to 4×10=2×10\sqrt{4 \times 10} = 2 \times \sqrt{10}.
  6. Substitute simplified square root: Substitute the simplified square root back into the solutions.\newlineFirst solution: x=−8−2⋅102=−8−10x = -8 - \frac{2 \cdot \sqrt{10}}{2} = -8 - \sqrt{10}.\newlineSecond solution: x=−8+2⋅102=−8+10x = -8 + \frac{2 \cdot \sqrt{10}}{2} = -8 + \sqrt{10}.
  7. Check solutions: Check if the solutions match any of the given answer choices.\newlineThe solutions we found are x=−8−10x = -8 - \sqrt{10} and x=−8+10x = -8 + \sqrt{10}, which match answer choice (A).

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