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X^2+2X-285=0
a = 1; b = 2; c = -285;
Δ = b2-4ac
Δ = 22-4·1·(-285)
Δ = 1144
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1144}=\sqrt{4*286}=\sqrt{4}*\sqrt{286}=2\sqrt{286}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{286}}{2*1}=\frac{-2-2\sqrt{286}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{286}}{2*1}=\frac{-2+2\sqrt{286}}{2} $
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