15x+4=20x^2+5x

Solution for 15x+4=20x^2+5x equation:

15x+4=20x^2+5x

We move all terms to the left:

15x+4-(20x^2+5x)=0

We get rid of parentheses

-20x^2+15x-5x+4=0

We add all the numbers together, and all the variables

-20x^2+10x+4=0

a = -20; b = 10; c = +4;
Δ = b2-4ac
Δ = 102-4·(-20)·4
Δ = 420
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{105}}{2*-20}=\frac{-10-2\sqrt{105}}{-40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{105}}{2*-20}=\frac{-10+2\sqrt{105}}{-40}
$