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20x^2+36x=-12x-16
We move all terms to the left:
20x^2+36x-(-12x-16)=0
We get rid of parentheses
20x^2+36x+12x+16=0
We add all the numbers together, and all the variables
20x^2+48x+16=0
a = 20; b = 48; c = +16;
Δ = b2-4ac
Δ = 482-4·20·16
Δ = 1024
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{1024}=32$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-32}{2*20}=\frac{-80}{40} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+32}{2*20}=\frac{-16}{40} =-2/5 $
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