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2x(3x+8)+(3x-4)=180
We move all terms to the left:
2x(3x+8)+(3x-4)-(180)=0
We multiply parentheses
6x^2+16x+(3x-4)-180=0
We get rid of parentheses
6x^2+16x+3x-4-180=0
We add all the numbers together, and all the variables
6x^2+19x-184=0
a = 6; b = 19; c = -184;
Δ = b2-4ac
Δ = 192-4·6·(-184)
Δ = 4777
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{4777}}{2*6}=\frac{-19-\sqrt{4777}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{4777}}{2*6}=\frac{-19+\sqrt{4777}}{12} $
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