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300=x(x+5)
We move all terms to the left:
300-(x(x+5))=0
We calculate terms in parentheses: -(x(x+5)), so:We get rid of parentheses
x(x+5)
We multiply parentheses
x^2+5x
Back to the equation:
-(x^2+5x)
-x^2-5x+300=0
We add all the numbers together, and all the variables
-1x^2-5x+300=0
a = -1; b = -5; c = +300;
Δ = b2-4ac
Δ = -52-4·(-1)·300
Δ = 1225
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{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-35}{2*-1}=\frac{-30}{-2} =+15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+35}{2*-1}=\frac{40}{-2} =-20 $
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