w^2+16w=27

Solution for w^2+16w=27 equation:

w^2+16w=27

We move all terms to the left:

w^2+16w-(27)=0

a = 1; b = 16; c = -27;
Δ = b2-4ac
Δ = 162-4·1·(-27)
Δ = 364
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:

$\sqrt{\Delta}=\sqrt{364}=\sqrt{4*91}=\sqrt{4}*\sqrt{91}=2\sqrt{91}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{91}}{2*1}=\frac{-16-2\sqrt{91}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{91}}{2*1}=\frac{-16+2\sqrt{91}}{2}
$