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2x^2+31x+42=0
a = 2; b = 31; c = +42;
Δ = b2-4ac
Δ = 312-4·2·42
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-25}{2*2}=\frac{-56}{4} =-14 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+25}{2*2}=\frac{-6}{4} =-1+1/2 $
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