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(5x+4)(x+4)=0
We multiply parentheses ..
(+5x^2+20x+4x+16)=0
We get rid of parentheses
5x^2+20x+4x+16=0
We add all the numbers together, and all the variables
5x^2+24x+16=0
a = 5; b = 24; c = +16;
Δ = b2-4ac
Δ = 242-4·5·16
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16}{2*5}=\frac{-40}{10} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16}{2*5}=\frac{-8}{10} =-4/5 $
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