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(2x-4)(5x+13)=0
We multiply parentheses ..
(+10x^2+26x-20x-52)=0
We get rid of parentheses
10x^2+26x-20x-52=0
We add all the numbers together, and all the variables
10x^2+6x-52=0
a = 10; b = 6; c = -52;
Δ = b2-4ac
Δ = 62-4·10·(-52)
Δ = 2116
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{2116}=46$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-46}{2*10}=\frac{-52}{20} =-2+3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+46}{2*10}=\frac{40}{20} =2 $
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