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.(5x+2)+(3x+1=)
We move all terms to the left:
.(5x+2)+(3x+1-())=0
We multiply parentheses
5x^2+(3x+1-())+=0
We calculate terms in parentheses: +(3x+1-()), so:We add all the numbers together, and all the variables
3x+1-()
We add all the numbers together, and all the variables
3x
Back to the equation:
+(3x)
5x^2+3x=0
a = 5; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·5·0
Δ = 9
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*5}=\frac{-6}{10} =-3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*5}=\frac{0}{10} =0 $
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