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2x^2+2x-1624=0
a = 2; b = 2; c = -1624;
Δ = b2-4ac
Δ = 22-4·2·(-1624)
Δ = 12996
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{12996}=114$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-114}{2*2}=\frac{-116}{4} =-29 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+114}{2*2}=\frac{112}{4} =28 $
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