12x^+28x=15x^2-24x

Solution for 12x^+28x=15x^2-24x equation:

12x^+28x=15x^2-24x

We move all terms to the left:

12x^+28x-(15x^2-24x)=0

We add all the numbers together, and all the variables

40x-(15x^2-24x)=0

We get rid of parentheses

-15x^2+40x+24x=0

We add all the numbers together, and all the variables

-15x^2+64x=0

a = -15; b = 64; c = 0;
Δ = b2-4ac
Δ = 642-4·(-15)·0
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-64}{2*-15}=\frac{-128}{-30}
=4+4/15
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+64}{2*-15}=\frac{0}{-30}
=0
$