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(3x+1)(x+3)=60
We move all terms to the left:
(3x+1)(x+3)-(60)=0
We multiply parentheses ..
(+3x^2+9x+x+3)-60=0
We get rid of parentheses
3x^2+9x+x+3-60=0
We add all the numbers together, and all the variables
3x^2+10x-57=0
a = 3; b = 10; c = -57;
Δ = b2-4ac
Δ = 102-4·3·(-57)
Δ = 784
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{784}=28$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-28}{2*3}=\frac{-38}{6} =-6+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+28}{2*3}=\frac{18}{6} =3 $
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