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(4.5(10^-4)x)-(.295x)=0
We add all the numbers together, and all the variables
(4.5(10^-4)x)-(+.295x)=0
We get rid of parentheses
(4.5(10^-4)x)-.295x=0
We calculate terms in parentheses: +(4.5(10^-4)x), so:We add all the numbers together, and all the variables
4.5(10^-4)x
We multiply parentheses
40x^2-16x
Back to the equation:
+(40x^2-16x)
-0.295x+(40x^2-16x)=0
We get rid of parentheses
40x^2-0.295x-16x=0
We add all the numbers together, and all the variables
40x^2-16.295x=0
a = 40; b = -16.295; c = 0;
Δ = b2-4ac
Δ = -16.2952-4·40·0
Δ = 265.527025
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16.295)-\sqrt{265.527025}}{2*40}=\frac{16.295-\sqrt{265.527025}}{80} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16.295)+\sqrt{265.527025}}{2*40}=\frac{16.295+\sqrt{265.527025}}{80} $
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