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(2x+1)(2x+2)=12
We move all terms to the left:
(2x+1)(2x+2)-(12)=0
We multiply parentheses ..
(+4x^2+4x+2x+2)-12=0
We get rid of parentheses
4x^2+4x+2x+2-12=0
We add all the numbers together, and all the variables
4x^2+6x-10=0
a = 4; b = 6; c = -10;
Δ = b2-4ac
Δ = 62-4·4·(-10)
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*4}=\frac{-20}{8} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*4}=\frac{8}{8} =1 $
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