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x^2+9x+13=3x+5
We move all terms to the left:
x^2+9x+13-(3x+5)=0
We get rid of parentheses
x^2+9x-3x-5+13=0
We add all the numbers together, and all the variables
x^2+6x+8=0
a = 1; b = 6; c = +8;
Δ = b2-4ac
Δ = 62-4·1·8
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*1}=\frac{-8}{2} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*1}=\frac{-4}{2} =-2 $
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