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(2x-5)(x=4)
We move all terms to the left:
(2x-5)(x-(4))=0
We multiply parentheses ..
(+2x^2-8x-5x+20)=0
We get rid of parentheses
2x^2-8x-5x+20=0
We add all the numbers together, and all the variables
2x^2-13x+20=0
a = 2; b = -13; c = +20;
Δ = b2-4ac
Δ = -132-4·2·20
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3}{2*2}=\frac{10}{4} =2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3}{2*2}=\frac{16}{4} =4 $
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