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x^2+2x=2400
We move all terms to the left:
x^2+2x-(2400)=0
a = 1; b = 2; c = -2400;
Δ = b2-4ac
Δ = 22-4·1·(-2400)
Δ = 9604
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{9604}=98$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-98}{2*1}=\frac{-100}{2} =-50 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+98}{2*1}=\frac{96}{2} =48 $
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