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4x^2+36x+84=4
We move all terms to the left:
4x^2+36x+84-(4)=0
We add all the numbers together, and all the variables
4x^2+36x+80=0
a = 4; b = 36; c = +80;
Δ = b2-4ac
Δ = 362-4·4·80
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4}{2*4}=\frac{-40}{8} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4}{2*4}=\frac{-32}{8} =-4 $
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