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(5x-3)(x+1)=0
We multiply parentheses ..
(+5x^2+5x-3x-3)=0
We get rid of parentheses
5x^2+5x-3x-3=0
We add all the numbers together, and all the variables
5x^2+2x-3=0
a = 5; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·5·(-3)
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*5}=\frac{-10}{10} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*5}=\frac{6}{10} =3/5 $
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