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2x^2+2x=7x+12
We move all terms to the left:
2x^2+2x-(7x+12)=0
We get rid of parentheses
2x^2+2x-7x-12=0
We add all the numbers together, and all the variables
2x^2-5x-12=0
a = 2; b = -5; c = -12;
Δ = b2-4ac
Δ = -52-4·2·(-12)
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*2}=\frac{-6}{4} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*2}=\frac{16}{4} =4 $
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