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2x^2-15x-23=0
a = 2; b = -15; c = -23;
Δ = b2-4ac
Δ = -152-4·2·(-23)
Δ = 409
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{409}}{2*2}=\frac{15-\sqrt{409}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{409}}{2*2}=\frac{15+\sqrt{409}}{4} $
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