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(x-3)(2x+5)+(x-3)(x-7)=0
We multiply parentheses ..
(+2x^2+5x-6x-15)+(x-3)(x-7)=0
We get rid of parentheses
2x^2+5x-6x+(x-3)(x-7)-15=0
We multiply parentheses ..
2x^2+(+x^2-7x-3x+21)+5x-6x-15=0
We add all the numbers together, and all the variables
2x^2+(+x^2-7x-3x+21)-1x-15=0
We get rid of parentheses
2x^2+x^2-7x-3x-1x+21-15=0
We add all the numbers together, and all the variables
3x^2-11x+6=0
a = 3; b = -11; c = +6;
Δ = b2-4ac
Δ = -112-4·3·6
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-7}{2*3}=\frac{4}{6} =2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+7}{2*3}=\frac{18}{6} =3 $
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