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3x+13=25+5x+5x+x^2
We move all terms to the left:
3x+13-(25+5x+5x+x^2)=0
We get rid of parentheses
-x^2-5x-5x+3x-25+13=0
We add all the numbers together, and all the variables
-1x^2-7x-12=0
a = -1; b = -7; c = -12;
Δ = b2-4ac
Δ = -72-4·(-1)·(-12)
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*-1}=\frac{6}{-2} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*-1}=\frac{8}{-2} =-4 $
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