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2x^2-5x-26=0
a = 2; b = -5; c = -26;
Δ = b2-4ac
Δ = -52-4·2·(-26)
Δ = 233
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{-(-5)-\sqrt{233}}{2*2}=\frac{5-\sqrt{233}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{233}}{2*2}=\frac{5+\sqrt{233}}{4} $
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