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2x^2+3x-6500=0
a = 2; b = 3; c = -6500;
Δ = b2-4ac
Δ = 32-4·2·(-6500)
Δ = 52009
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{-(3)-\sqrt{52009}}{2*2}=\frac{-3-\sqrt{52009}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{52009}}{2*2}=\frac{-3+\sqrt{52009}}{4} $
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