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2x^2-23x-25=0
a = 2; b = -23; c = -25;
Δ = b2-4ac
Δ = -232-4·2·(-25)
Δ = 729
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{729}=27$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-27}{2*2}=\frac{-4}{4} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+27}{2*2}=\frac{50}{4} =12+1/2 $
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