The x-intercetps of the parabola y = 2x^2 + 13x - 4\(+ 2x^2 - 6x + 5\) are at (p,0) and (q,0). Find p and q. Enter your answer in the form "p,q".
Find p and q.
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\( y = 2x^2 + 13x - 4+ 2x^2 - 6x + 5\\ y=4x^2+7x+1\\ y=x^2+1.75x+0.25=0 \)
\(x=-0.875\pm \sqrt{0.515625}\)
\(\{p,q\}=\{-1.59307033082,-0.156929669183\}\)
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