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(3x-5)+(19-x)2x=
We move all terms to the left:
(3x-5)+(19-x)2x-()=0
We add all the numbers together, and all the variables
(3x-5)+(-1x+19)2x-()=0
We add all the numbers together, and all the variables
(3x-5)+(-1x+19)2x=0
We multiply parentheses
-2x^2+(3x-5)+38x=0
We get rid of parentheses
-2x^2+3x+38x-5=0
We add all the numbers together, and all the variables
-2x^2+41x-5=0
a = -2; b = 41; c = -5;
Δ = b2-4ac
Δ = 412-4·(-2)·(-5)
Δ = 1641
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{-(41)-\sqrt{1641}}{2*-2}=\frac{-41-\sqrt{1641}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+\sqrt{1641}}{2*-2}=\frac{-41+\sqrt{1641}}{-4} $
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