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(2x+9)(4x+3)=120
We move all terms to the left:
(2x+9)(4x+3)-(120)=0
We multiply parentheses ..
(+8x^2+6x+36x+27)-120=0
We get rid of parentheses
8x^2+6x+36x+27-120=0
We add all the numbers together, and all the variables
8x^2+42x-93=0
a = 8; b = 42; c = -93;
Δ = b2-4ac
Δ = 422-4·8·(-93)
Δ = 4740
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4740}=\sqrt{4*1185}=\sqrt{4}*\sqrt{1185}=2\sqrt{1185}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{1185}}{2*8}=\frac{-42-2\sqrt{1185}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{1185}}{2*8}=\frac{-42+2\sqrt{1185}}{16} $
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