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(x+1)(x+3)+(x+3)(4x-5)=0
We multiply parentheses ..
(+x^2+3x+x+3)+(x+3)(4x-5)=0
We get rid of parentheses
x^2+3x+x+(x+3)(4x-5)+3=0
We multiply parentheses ..
x^2+(+4x^2-5x+12x-15)+3x+x+3=0
We add all the numbers together, and all the variables
x^2+(+4x^2-5x+12x-15)+4x+3=0
We get rid of parentheses
x^2+4x^2-5x+12x+4x-15+3=0
We add all the numbers together, and all the variables
5x^2+11x-12=0
a = 5; b = 11; c = -12;
Δ = b2-4ac
Δ = 112-4·5·(-12)
Δ = 361
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{361}=19$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*5}=\frac{-30}{10} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*5}=\frac{8}{10} =4/5 $
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