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3x(x+2)=-2(-x-5)
We move all terms to the left:
3x(x+2)-(-2(-x-5))=0
We add all the numbers together, and all the variables
3x(x+2)-(-2(-1x-5))=0
We multiply parentheses
3x^2+6x-(-2(-1x-5))=0
We calculate terms in parentheses: -(-2(-1x-5)), so:We get rid of parentheses
-2(-1x-5)
We multiply parentheses
2x+10
Back to the equation:
-(2x+10)
3x^2+6x-2x-10=0
We add all the numbers together, and all the variables
3x^2+4x-10=0
a = 3; b = 4; c = -10;
Δ = b2-4ac
Δ = 42-4·3·(-10)
Δ = 136
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{34}}{2*3}=\frac{-4-2\sqrt{34}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{34}}{2*3}=\frac{-4+2\sqrt{34}}{6} $
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