4x^2+12x=41

Solution for 4x^2+12x=41 equation:

4x^2+12x=41

We move all terms to the left:

4x^2+12x-(41)=0

a = 4; b = 12; c = -41;
Δ = b2-4ac
Δ = 122-4·4·(-41)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-20\sqrt{2}}{2*4}=\frac{-12-20\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+20\sqrt{2}}{2*4}=\frac{-12+20\sqrt{2}}{8}
$