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3x^2=2x+12-7x
We move all terms to the left:
3x^2-(2x+12-7x)=0
We add all the numbers together, and all the variables
3x^2-(-5x+12)=0
We get rid of parentheses
3x^2+5x-12=0
a = 3; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·3·(-12)
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*3}=\frac{-18}{6} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*3}=\frac{8}{6} =1+1/3 $
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