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(3x+2)(x+2)=35
We move all terms to the left:
(3x+2)(x+2)-(35)=0
We multiply parentheses ..
(+3x^2+6x+2x+4)-35=0
We get rid of parentheses
3x^2+6x+2x+4-35=0
We add all the numbers together, and all the variables
3x^2+8x-31=0
a = 3; b = 8; c = -31;
Δ = b2-4ac
Δ = 82-4·3·(-31)
Δ = 436
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{109}}{2*3}=\frac{-8-2\sqrt{109}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{109}}{2*3}=\frac{-8+2\sqrt{109}}{6} $
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