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2x^2-15x-22=0
a = 2; b = -15; c = -22;
Δ = b2-4ac
Δ = -152-4·2·(-22)
Δ = 401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{401}}{2*2}=\frac{15-\sqrt{401}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{401}}{2*2}=\frac{15+\sqrt{401}}{4} $
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