2x+5=4x^2+2x

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

2x+5=4x^2+2x

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4x^2+5=0

a = -4; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-4)·5
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-4}=\frac{0-4\sqrt{5}}{-8}
=-\frac{4\sqrt{5}}{-8}
=-\frac{\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-4}=\frac{0+4\sqrt{5}}{-8}
=\frac{4\sqrt{5}}{-8}
=\frac{\sqrt{5}}{-2}
$