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10w^2+5w-63=0
a = 10; b = 5; c = -63;
Δ = b2-4ac
Δ = 52-4·10·(-63)
Δ = 2545
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{2545}}{2*10}=\frac{-5-\sqrt{2545}}{20} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{2545}}{2*10}=\frac{-5+\sqrt{2545}}{20} $
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