2x+5=4x^2

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

2x+5=4x^2

We move all terms to the left:

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

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

a = -4; b = 2; c = +5;
Δ = b2-4ac
Δ = 22-4·(-4)·5
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{21}}{2*-4}=\frac{-2-2\sqrt{21}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{21}}{2*-4}=\frac{-2+2\sqrt{21}}{-8}
$