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(2d-5)(2d-4)=4d+d=20
We move all terms to the left:
(2d-5)(2d-4)-(4d+d)=0
We add all the numbers together, and all the variables
(2d-5)(2d-4)-(+5d)=0
We get rid of parentheses
(2d-5)(2d-4)-5d=0
We multiply parentheses ..
(+4d^2-8d-10d+20)-5d=0
We get rid of parentheses
4d^2-8d-10d-5d+20=0
We add all the numbers together, and all the variables
4d^2-23d+20=0
a = 4; b = -23; c = +20;
Δ = b2-4ac
Δ = -232-4·4·20
Δ = 209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{209}}{2*4}=\frac{23-\sqrt{209}}{8} $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{209}}{2*4}=\frac{23+\sqrt{209}}{8} $
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