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x^2+5x+36+10x=360
We move all terms to the left:
x^2+5x+36+10x-(360)=0
We add all the numbers together, and all the variables
x^2+15x-324=0
a = 1; b = 15; c = -324;
Δ = b2-4ac
Δ = 152-4·1·(-324)
Δ = 1521
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{1521}=39$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-39}{2*1}=\frac{-54}{2} =-27 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+39}{2*1}=\frac{24}{2} =12 $
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