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2x^2-25x-42=0
a = 2; b = -25; c = -42;
Δ = b2-4ac
Δ = -252-4·2·(-42)
Δ = 961
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{961}=31$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-31}{2*2}=\frac{-6}{4} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+31}{2*2}=\frac{56}{4} =14 $
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