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5x(2x+6)=-4(-5-2x)+3x
We move all terms to the left:
5x(2x+6)-(-4(-5-2x)+3x)=0
We add all the numbers together, and all the variables
5x(2x+6)-(-4(-2x-5)+3x)=0
We multiply parentheses
10x^2+30x-(-4(-2x-5)+3x)=0
We calculate terms in parentheses: -(-4(-2x-5)+3x), so:We get rid of parentheses
-4(-2x-5)+3x
We add all the numbers together, and all the variables
3x-4(-2x-5)
We multiply parentheses
3x+8x+20
We add all the numbers together, and all the variables
11x+20
Back to the equation:
-(11x+20)
10x^2+30x-11x-20=0
We add all the numbers together, and all the variables
10x^2+19x-20=0
a = 10; b = 19; c = -20;
Δ = b2-4ac
Δ = 192-4·10·(-20)
Δ = 1161
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1161}=\sqrt{9*129}=\sqrt{9}*\sqrt{129}=3\sqrt{129}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-3\sqrt{129}}{2*10}=\frac{-19-3\sqrt{129}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+3\sqrt{129}}{2*10}=\frac{-19+3\sqrt{129}}{20} $
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