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12x^2+31x+13=-7
We move all terms to the left:
12x^2+31x+13-(-7)=0
We add all the numbers together, and all the variables
12x^2+31x+20=0
a = 12; b = 31; c = +20;
Δ = b2-4ac
Δ = 312-4·12·20
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-1}{2*12}=\frac{-32}{24} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+1}{2*12}=\frac{-30}{24} =-1+1/4 $
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