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Use the quadratic formula to solve. Express your answer in simplest form. 8t^(2)-19 t+12=4t^(2) Answer: t=

Use the quadratic formula to solve. Express your answer in simplest form.\newline8t2−19t+12=4t2 8 t^{2}-19 t+12=4 t^{2} \newlineAnswer: t= t=

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Q. Use the quadratic formula to solve. Express your answer in simplest form.\newline8t2−19t+12=4t2 8 t^{2}-19 t+12=4 t^{2} \newlineAnswer: t= t=
  1. Set Equation to Zero: First, we need to set the equation to zero by subtracting 4t24t^2 from both sides of the equation.\newline8t2−19t+12−4t2=08t^2 - 19t + 12 - 4t^2 = 0\newlineThis simplifies to:\newline4t2−19t+12=04t^2 - 19t + 12 = 0
  2. Apply Quadratic Formula: Now we can apply the quadratic formula to solve for tt. The quadratic formula is given by:\newlinet=−b±b2−4ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\newlinewhere aa, bb, and cc are the coefficients from the quadratic equation at2+bt+c=0at^2 + bt + c = 0.\newlineIn our equation, a=4a = 4, b=−19b = -19, and c=12c = 12.
  3. Calculate Discriminant: Next, we calculate the discriminant, which is the part under the square root in the quadratic formula: b2−4acb^2 - 4ac.\newlineDiscriminant = (−19)2−4(4)(12)(-19)^2 - 4(4)(12)\newlineDiscriminant = 361−192361 - 192\newlineDiscriminant = 169169
  4. Find Solutions: Since the discriminant is positive, we will have two real solutions. We can now plug the values of aa, bb, and the discriminant into the quadratic formula to find the solutions for tt.
    t=−(−19)±1692×4t = \frac{-(-19) \pm \sqrt{169}}{2 \times 4}
    t=19±138t = \frac{19 \pm 13}{8}
  5. Two Real Solutions: We will have two solutions, one for the addition and one for the subtraction:\newlinet1=19+138t_1 = \frac{19 + 13}{8}\newlinet1=328t_1 = \frac{32}{8}\newlinet1=4t_1 = 4\newlinet2=19−138t_2 = \frac{19 - 13}{8}\newlinet2=68t_2 = \frac{6}{8}\newlinet2=34t_2 = \frac{3}{4}

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