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3x(2x+5)=3x/4+36
We move all terms to the left:
3x(2x+5)-(3x/4+36)=0
We multiply parentheses
6x^2+15x-(3x/4+36)=0
We get rid of parentheses
6x^2+15x-3x/4-36=0
We multiply all the terms by the denominator
6x^2*4+15x*4-3x-36*4=0
We add all the numbers together, and all the variables
6x^2*4-3x+15x*4-144=0
Wy multiply elements
24x^2-3x+60x-144=0
We add all the numbers together, and all the variables
24x^2+57x-144=0
a = 24; b = 57; c = -144;
Δ = b2-4ac
Δ = 572-4·24·(-144)
Δ = 17073
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{17073}=\sqrt{9*1897}=\sqrt{9}*\sqrt{1897}=3\sqrt{1897}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(57)-3\sqrt{1897}}{2*24}=\frac{-57-3\sqrt{1897}}{48} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(57)+3\sqrt{1897}}{2*24}=\frac{-57+3\sqrt{1897}}{48} $
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