Question

Given y= 2x squared - 10x find all the real values of x for which y = -3

Answer 1

`y=2 x^(2) -10x \\ -3=2 x^(2) -10x \\ 2 x^(2) -10x+3=0 \\ x=\frac{-b \pm \sqrt{ b^(2) -4ac} }{2a}, \ where \ a=2, \ b=-10, \ c=3`
Therefore,

`x=\frac{-(-10) \pm \sqrt{ (-10)^(2) -(4 * 2 * 3)} }{2 * 2} \\ =(10 \pm √( 100 -24) )/(4) \\ =(10 \pm √( 76) )/(4) \\= (10 + √( 76) )/(4) \ or \ (10 - √( 76) )/(4) \\ =4.679 \ or \ 0.3206`