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2w^2+16w=36
We move all terms to the left:
2w^2+16w-(36)=0
a = 2; b = 16; c = -36;
Δ = b2-4ac
Δ = 162-4·2·(-36)
Δ = 544
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{34}}{2*2}=\frac{-16-4\sqrt{34}}{4} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{34}}{2*2}=\frac{-16+4\sqrt{34}}{4} $
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