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16x^2+31x-15=0
a = 16; b = 31; c = -15;
Δ = b2-4ac
Δ = 312-4·16·(-15)
Δ = 1921
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{1921}}{2*16}=\frac{-31-\sqrt{1921}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+\sqrt{1921}}{2*16}=\frac{-31+\sqrt{1921}}{32} $
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