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(4x+5)(x+4)=0
We multiply parentheses ..
(+4x^2+16x+5x+20)=0
We get rid of parentheses
4x^2+16x+5x+20=0
We add all the numbers together, and all the variables
4x^2+21x+20=0
a = 4; b = 21; c = +20;
Δ = b2-4ac
Δ = 212-4·4·20
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*4}=\frac{-32}{8} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*4}=\frac{-10}{8} =-1+1/4 $
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