8x^2=-48+40x

Solution for 8x^2=-48+40x equation:

8x^2=-48+40x

We move all terms to the left:

8x^2-(-48+40x)=0

We add all the numbers together, and all the variables

8x^2-(40x-48)=0

We get rid of parentheses

8x^2-40x+48=0

a = 8; b = -40; c = +48;
Δ = b2-4ac
Δ = -402-4·8·48
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{-(-40)-8}{2*8}=\frac{32}{16}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8}{2*8}=\frac{48}{16}
=3
$