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(2x+3)(2x+3)+(2x+3)(7x-2)=0
We multiply parentheses ..
(+4x^2+6x+6x+9)+(2x+3)(7x-2)=0
We get rid of parentheses
4x^2+6x+6x+(2x+3)(7x-2)+9=0
We multiply parentheses ..
4x^2+(+14x^2-4x+21x-6)+6x+6x+9=0
We add all the numbers together, and all the variables
4x^2+(+14x^2-4x+21x-6)+12x+9=0
We get rid of parentheses
4x^2+14x^2-4x+21x+12x-6+9=0
We add all the numbers together, and all the variables
18x^2+29x+3=0
a = 18; b = 29; c = +3;
Δ = b2-4ac
Δ = 292-4·18·3
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-25}{2*18}=\frac{-54}{36} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+25}{2*18}=\frac{-4}{36} =-1/9 $
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