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(3x+2)(x+5)=120
We move all terms to the left:
(3x+2)(x+5)-(120)=0
We multiply parentheses ..
(+3x^2+15x+2x+10)-120=0
We get rid of parentheses
3x^2+15x+2x+10-120=0
We add all the numbers together, and all the variables
3x^2+17x-110=0
a = 3; b = 17; c = -110;
Δ = b2-4ac
Δ = 172-4·3·(-110)
Δ = 1609
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{-(17)-\sqrt{1609}}{2*3}=\frac{-17-\sqrt{1609}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{1609}}{2*3}=\frac{-17+\sqrt{1609}}{6} $
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