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(x-5)(5x+1)+(5x+1)(x+10)=0
We multiply parentheses ..
(+5x^2+x-25x-5)+(5x+1)(x+10)=0
We get rid of parentheses
5x^2+x-25x+(5x+1)(x+10)-5=0
We multiply parentheses ..
5x^2+(+5x^2+50x+x+10)+x-25x-5=0
We add all the numbers together, and all the variables
5x^2+(+5x^2+50x+x+10)-24x-5=0
We get rid of parentheses
5x^2+5x^2+50x+x-24x+10-5=0
We add all the numbers together, and all the variables
10x^2+27x+5=0
a = 10; b = 27; c = +5;
Δ = b2-4ac
Δ = 272-4·10·5
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-23}{2*10}=\frac{-50}{20} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+23}{2*10}=\frac{-4}{20} =-1/5 $
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