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(3x-7)(5+2x)=11
We move all terms to the left:
(3x-7)(5+2x)-(11)=0
We add all the numbers together, and all the variables
(3x-7)(2x+5)-11=0
We multiply parentheses ..
(+6x^2+15x-14x-35)-11=0
We get rid of parentheses
6x^2+15x-14x-35-11=0
We add all the numbers together, and all the variables
6x^2+x-46=0
a = 6; b = 1; c = -46;
Δ = b2-4ac
Δ = 12-4·6·(-46)
Δ = 1105
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{-(1)-\sqrt{1105}}{2*6}=\frac{-1-\sqrt{1105}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1105}}{2*6}=\frac{-1+\sqrt{1105}}{12} $
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