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(5x-3)(2x+1)=4x+5
We move all terms to the left:
(5x-3)(2x+1)-(4x+5)=0
We get rid of parentheses
(5x-3)(2x+1)-4x-5=0
We multiply parentheses ..
(+10x^2+5x-6x-3)-4x-5=0
We get rid of parentheses
10x^2+5x-6x-4x-3-5=0
We add all the numbers together, and all the variables
10x^2-5x-8=0
a = 10; b = -5; c = -8;
Δ = b2-4ac
Δ = -52-4·10·(-8)
Δ = 345
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{345}}{2*10}=\frac{5-\sqrt{345}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{345}}{2*10}=\frac{5+\sqrt{345}}{20} $
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