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4x^2+12x=135
We move all terms to the left:
4x^2+12x-(135)=0
a = 4; b = 12; c = -135;
Δ = b2-4ac
Δ = 122-4·4·(-135)
Δ = 2304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-48}{2*4}=\frac{-60}{8} =-7+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+48}{2*4}=\frac{36}{8} =4+1/2 $
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