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(2x+1)(x+4)+(x+4)(3+-5x)=0
We add all the numbers together, and all the variables
(2x+1)(x+4)+(x+4)(-5x)=0
We multiply parentheses ..
(+2x^2+8x+x+4)+(x+4)(-5x)=0
We get rid of parentheses
2x^2+8x+x+(x+4)(-5x)+4=0
We multiply parentheses ..
2x^2+(-5x^2-20x)+8x+x+4=0
We add all the numbers together, and all the variables
2x^2+(-5x^2-20x)+9x+4=0
We get rid of parentheses
2x^2-5x^2-20x+9x+4=0
We add all the numbers together, and all the variables
-3x^2-11x+4=0
a = -3; b = -11; c = +4;
Δ = b2-4ac
Δ = -112-4·(-3)·4
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*-3}=\frac{-2}{-6} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*-3}=\frac{24}{-6} =-4 $
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