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x^2+4x=306
We move all terms to the left:
x^2+4x-(306)=0
a = 1; b = 4; c = -306;
Δ = b2-4ac
Δ = 42-4·1·(-306)
Δ = 1240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1240}=\sqrt{4*310}=\sqrt{4}*\sqrt{310}=2\sqrt{310}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{310}}{2*1}=\frac{-4-2\sqrt{310}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{310}}{2*1}=\frac{-4+2\sqrt{310}}{2} $
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