Question

Just need to solve number 8 with quadratic formula

Answer 1

`ax^2+bx+c=0\\\\x_1=(-b-√(b^2-4ac))/(2a)\\\\x_2=(-b+√(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\\\\√(b^2-4ac)=\sqrt8=√(4\cdot2)=2\sqrt2`

`x_1=(-(-8)-2\sqrt2)/(2\cdot2)=(4-\sqrt2)/(2)\\\\x_2=(-(-8)+2\sqrt2)/(2\cdot2)=(4+\sqrt2)/(2)`