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Find one value of x that is a solution to the equation: {:[(5x+2)^(2)+15 x+6=0],[x=◻]:}

Find one value of x x that is a solution to the equation:\newline(5x+2)2+15x+6=0x=□ \begin{array}{l} (5 x+2)^{2}+15 x+6=0 \\ x=\square \end{array}

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Full solution

Q. Find one value of x x that is a solution to the equation:\newline(5x+2)2+15x+6=0x=□ \begin{array}{l} (5 x+2)^{2}+15 x+6=0 \\ x=\square \end{array}
  1. Write equation: Write down the given equation.\newlineWe are given the equation (5x+2)2+15x+6=0(5x+2)^2 + 15x + 6 = 0.
  2. Expand squared term: Expand the squared term.\newline(5x+2)2=(5x+2)(5x+2)=25x2+10x+10x+4=25x2+20x+4(5x+2)^2 = (5x+2)(5x+2) = 25x^2 + 10x + 10x + 4 = 25x^2 + 20x + 4\newlineNow, substitute this back into the original equation.\newline25x2+20x+4+15x+6=025x^2 + 20x + 4 + 15x + 6 = 0
  3. Substitute back: Combine like terms.\newline25x2+20x+15x+4+6=025x^2 + 20x + 15x + 4 + 6 = 0\newline25x2+35x+10=025x^2 + 35x + 10 = 0
  4. Combine like terms: Factor the quadratic equation.\newlineWe need to find two numbers that multiply to 25×10=25025 \times 10 = 250 and add up to 3535.\newlineThese numbers are 2525 and 1010.\newlineSo we can write the equation as:\newline(5x+5)(5x+2)=0(5x + 5)(5x + 2) = 0
  5. Factor quadratic equation: Solve for x by setting each factor equal to zero.\newlineFirst factor: 5x+5=05x + 5 = 0\newline5x=−55x = -5\newlinex=−55x = \frac{-5}{5}\newlinex=−1x = -1\newlineSecond factor: 5x+2=05x + 2 = 0\newline5x=−25x = -2\newlinex=−25x = \frac{-2}{5}\newline$x = \(-0\).\(4\)

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