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(3x-1)(2x+5)=5
We move all terms to the left:
(3x-1)(2x+5)-(5)=0
We multiply parentheses ..
(+6x^2+15x-2x-5)-5=0
We get rid of parentheses
6x^2+15x-2x-5-5=0
We add all the numbers together, and all the variables
6x^2+13x-10=0
a = 6; b = 13; c = -10;
Δ = b2-4ac
Δ = 132-4·6·(-10)
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{-(13)-\sqrt{409}}{2*6}=\frac{-13-\sqrt{409}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{409}}{2*6}=\frac{-13+\sqrt{409}}{12} $
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