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2x^2+12x-589=0
a = 2; b = 12; c = -589;
Δ = b2-4ac
Δ = 122-4·2·(-589)
Δ = 4856
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4856}=\sqrt{4*1214}=\sqrt{4}*\sqrt{1214}=2\sqrt{1214}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{1214}}{2*2}=\frac{-12-2\sqrt{1214}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{1214}}{2*2}=\frac{-12+2\sqrt{1214}}{4} $
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