Question

Just need to solve number 8 with quadratic formula

Answer 1

$ax^2+bx+c=0\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\\\\x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}$

$2x^2=8x-7\ \ \ \ |-8x\\\\2x^2-8x=-7\ \ \ |+7\\\\2x^2-8x+7=0\\\\a=2;\ b=-8;\ c=7\\\\b^2-4ac=(-8)^2-4\cdot2\cdot7=64-56=8\\\\\sqrt{b^2-4ac}=\sqrt8=\sqrt{4\cdot2}=2\sqrt2$
$x_1=\dfrac{-(-8)-2\sqrt2}{2\cdot2}=\dfrac{4-\sqrt2}{2}\\\\x_2=\dfrac{-(-8)+2\sqrt2}{2\cdot2}=\dfrac{4+\sqrt2}{2}$