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w^2+4w-40=0
a = 1; b = 4; c = -40;
Δ = b2-4ac
Δ = 42-4·1·(-40)
Δ = 176
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{11}}{2*1}=\frac{-4-4\sqrt{11}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{11}}{2*1}=\frac{-4+4\sqrt{11}}{2} $
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