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(3x+4)(2x+1)=26
We move all terms to the left:
(3x+4)(2x+1)-(26)=0
We multiply parentheses ..
(+6x^2+3x+8x+4)-26=0
We get rid of parentheses
6x^2+3x+8x+4-26=0
We add all the numbers together, and all the variables
6x^2+11x-22=0
a = 6; b = 11; c = -22;
Δ = b2-4ac
Δ = 112-4·6·(-22)
Δ = 649
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{649}}{2*6}=\frac{-11-\sqrt{649}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{649}}{2*6}=\frac{-11+\sqrt{649}}{12} $
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