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(3x+8)(2x+13)=160
We move all terms to the left:
(3x+8)(2x+13)-(160)=0
We multiply parentheses ..
(+6x^2+39x+16x+104)-160=0
We get rid of parentheses
6x^2+39x+16x+104-160=0
We add all the numbers together, and all the variables
6x^2+55x-56=0
a = 6; b = 55; c = -56;
Δ = b2-4ac
Δ = 552-4·6·(-56)
Δ = 4369
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{-(55)-\sqrt{4369}}{2*6}=\frac{-55-\sqrt{4369}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+\sqrt{4369}}{2*6}=\frac{-55+\sqrt{4369}}{12} $
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