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w^2+15w-24=0
a = 1; b = 15; c = -24;
Δ = b2-4ac
Δ = 152-4·1·(-24)
Δ = 321
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{321}}{2*1}=\frac{-15-\sqrt{321}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{321}}{2*1}=\frac{-15+\sqrt{321}}{2} $
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