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3x(x-4)+15=2(x+3)
We move all terms to the left:
3x(x-4)+15-(2(x+3))=0
We multiply parentheses
3x^2-12x-(2(x+3))+15=0
We calculate terms in parentheses: -(2(x+3)), so:We get rid of parentheses
2(x+3)
We multiply parentheses
2x+6
Back to the equation:
-(2x+6)
3x^2-12x-2x-6+15=0
We add all the numbers together, and all the variables
3x^2-14x+9=0
a = 3; b = -14; c = +9;
Δ = b2-4ac
Δ = -142-4·3·9
Δ = 88
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{22}}{2*3}=\frac{14-2\sqrt{22}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{22}}{2*3}=\frac{14+2\sqrt{22}}{6} $
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