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5x^2+31x+20=0
a = 5; b = 31; c = +20;
Δ = b2-4ac
Δ = 312-4·5·20
Δ = 561
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{-(31)-\sqrt{561}}{2*5}=\frac{-31-\sqrt{561}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+\sqrt{561}}{2*5}=\frac{-31+\sqrt{561}}{10} $
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