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1=5x^2+4x
We move all terms to the left:
1-(5x^2+4x)=0
We get rid of parentheses
-5x^2-4x+1=0
a = -5; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·(-5)·1
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*-5}=\frac{-2}{-10} =1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*-5}=\frac{10}{-10} =-1 $
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