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x2-12=-2x^2-16x
We move all terms to the left:
x2-12-(-2x^2-16x)=0
We add all the numbers together, and all the variables
x^2-(-2x^2-16x)-12=0
We get rid of parentheses
x^2+2x^2+16x-12=0
We add all the numbers together, and all the variables
3x^2+16x-12=0
a = 3; b = 16; c = -12;
Δ = b2-4ac
Δ = 162-4·3·(-12)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-20}{2*3}=\frac{-36}{6} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+20}{2*3}=\frac{4}{6} =2/3 $
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