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X^2+3X=364
We move all terms to the left:
X^2+3X-(364)=0
a = 1; b = 3; c = -364;
Δ = b2-4ac
Δ = 32-4·1·(-364)
Δ = 1465
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{1465}}{2*1}=\frac{-3-\sqrt{1465}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{1465}}{2*1}=\frac{-3+\sqrt{1465}}{2} $
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