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3x(x+5)-3=2(3x*1)-3x
We move all terms to the left:
3x(x+5)-3-(2(3x*1)-3x)=0
We add all the numbers together, and all the variables
3x(x+5)-(2(+3x*1)-3x)-3=0
We multiply parentheses
3x^2+15x-(2(+3x*1)-3x)-3=0
We calculate terms in parentheses: -(2(+3x*1)-3x), so:We add all the numbers together, and all the variables
2(+3x*1)-3x
We add all the numbers together, and all the variables
-3x+2(+3x*1)
We multiply parentheses
-3x+6x
We add all the numbers together, and all the variables
3x
Back to the equation:
-(3x)
3x^2+12x-3=0
a = 3; b = 12; c = -3;
Δ = b2-4ac
Δ = 122-4·3·(-3)
Δ = 180
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{5}}{2*3}=\frac{-12-6\sqrt{5}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{5}}{2*3}=\frac{-12+6\sqrt{5}}{6} $
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