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(x+1)(x+3)+(x+3)(4x-5)=342
We move all terms to the left:
(x+1)(x+3)+(x+3)(4x-5)-(342)=0
We multiply parentheses ..
(+x^2+3x+x+3)+(x+3)(4x-5)-342=0
We get rid of parentheses
x^2+3x+x+(x+3)(4x-5)+3-342=0
We multiply parentheses ..
x^2+(+4x^2-5x+12x-15)+3x+x+3-342=0
We add all the numbers together, and all the variables
x^2+(+4x^2-5x+12x-15)+4x-339=0
We get rid of parentheses
x^2+4x^2-5x+12x+4x-15-339=0
We add all the numbers together, and all the variables
5x^2+11x-354=0
a = 5; b = 11; c = -354;
Δ = b2-4ac
Δ = 112-4·5·(-354)
Δ = 7201
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{7201}}{2*5}=\frac{-11-\sqrt{7201}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{7201}}{2*5}=\frac{-11+\sqrt{7201}}{10} $
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