3w^2+22w=7=0

Solution for 3w^2+22w=7=0 equation:

3w^2+22w=7=0

We move all terms to the left:

3w^2+22w-(7)=0

a = 3; b = 22; c = -7;
Δ = b2-4ac
Δ = 222-4·3·(-7)
Δ = 568
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{568}=\sqrt{4*142}=\sqrt{4}*\sqrt{142}=2\sqrt{142}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{142}}{2*3}=\frac{-22-2\sqrt{142}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{142}}{2*3}=\frac{-22+2\sqrt{142}}{6}
$