-2x^2+20x=48

Solution for -2x^2+20x=48 equation:

-2x^2+20x=48

We move all terms to the left:

-2x^2+20x-(48)=0

a = -2; b = 20; c = -48;
Δ = b2-4ac
Δ = 202-4·(-2)·(-48)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*-2}=\frac{-24}{-4}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*-2}=\frac{-16}{-4}
=+4
$