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3x(2x+5)=2(2x+6)
We move all terms to the left:
3x(2x+5)-(2(2x+6))=0
We multiply parentheses
6x^2+15x-(2(2x+6))=0
We calculate terms in parentheses: -(2(2x+6)), so:We get rid of parentheses
2(2x+6)
We multiply parentheses
4x+12
Back to the equation:
-(4x+12)
6x^2+15x-4x-12=0
We add all the numbers together, and all the variables
6x^2+11x-12=0
a = 6; b = 11; c = -12;
Δ = b2-4ac
Δ = 112-4·6·(-12)
Δ = 409
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{-(11)-\sqrt{409}}{2*6}=\frac{-11-\sqrt{409}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{409}}{2*6}=\frac{-11+\sqrt{409}}{12} $
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