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12=4x+5x^2
We move all terms to the left:
12-(4x+5x^2)=0
We get rid of parentheses
-5x^2-4x+12=0
a = -5; b = -4; c = +12;
Δ = b2-4ac
Δ = -42-4·(-5)·12
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-16}{2*-5}=\frac{-12}{-10} =1+1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*-5}=\frac{20}{-10} =-2 $
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