4u2+4u=15

Solution for 4u2+4u=15 equation:

4u^2+4u=15

We move all terms to the left:

4u^2+4u-(15)=0

a = 4; b = 4; c = -15;
Δ = b2-4ac
Δ = 42-4·4·(-15)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*4}=\frac{-20}{8}
=-2+1/2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*4}=\frac{12}{8}
=1+1/2
$