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(10+x)(3+x)=60
We move all terms to the left:
(10+x)(3+x)-(60)=0
We add all the numbers together, and all the variables
(x+10)(x+3)-60=0
We multiply parentheses ..
(+x^2+3x+10x+30)-60=0
We get rid of parentheses
x^2+3x+10x+30-60=0
We add all the numbers together, and all the variables
x^2+13x-30=0
a = 1; b = 13; c = -30;
Δ = b2-4ac
Δ = 132-4·1·(-30)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*1}=\frac{-30}{2} =-15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*1}=\frac{4}{2} =2 $
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