-2x^2+20x=42

Solution for -2x^2+20x=42 equation:

-2x^2+20x=42

We move all terms to the left:

-2x^2+20x-(42)=0

a = -2; b = 20; c = -42;
Δ = b2-4ac
Δ = 202-4·(-2)·(-42)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8}{2*-2}=\frac{-28}{-4}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8}{2*-2}=\frac{-12}{-4}
=+3
$