4x^2=4x+21

Solution for 4x^2=4x+21 equation:

4x^2=4x+21

We move all terms to the left:

4x^2-(4x+21)=0

We get rid of parentheses

4x^2-4x-21=0

a = 4; b = -4; c = -21;
Δ = b2-4ac
Δ = -42-4·4·(-21)
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{22}}{2*4}=\frac{4-4\sqrt{22}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{22}}{2*4}=\frac{4+4\sqrt{22}}{8}
$