40x^2=-8+48x

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

40x^2=-8+48x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

40x^2-48x+8=0

a = 40; b = -48; c = +8;
Δ = b2-4ac
Δ = -482-4·40·8
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-32}{2*40}=\frac{16}{80}
=1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+32}{2*40}=\frac{80}{80}
=1
$