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4w^2-5w-90=0
a = 4; b = -5; c = -90;
Δ = b2-4ac
Δ = -52-4·4·(-90)
Δ = 1465
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{-(-5)-\sqrt{1465}}{2*4}=\frac{5-\sqrt{1465}}{8} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{1465}}{2*4}=\frac{5+\sqrt{1465}}{8} $
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