22x^2+4x=840

Solution for 22x^2+4x=840 equation:

22x^2+4x=840

We move all terms to the left:

22x^2+4x-(840)=0

a = 22; b = 4; c = -840;
Δ = b2-4ac
Δ = 42-4·22·(-840)
Δ = 73936
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:

$\sqrt{\Delta}=\sqrt{73936}=\sqrt{16*4621}=\sqrt{16}*\sqrt{4621}=4\sqrt{4621}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{4621}}{2*22}=\frac{-4-4\sqrt{4621}}{44}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{4621}}{2*22}=\frac{-4+4\sqrt{4621}}{44}
$