(5/4)(20x+12)=-35

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Solution for (5/4)(20x+12)=-35 equation: 



(5/4)(20x+12)=-35

We move all terms to the left:

(5/4)(20x+12)-(-35)=0



Domain of the equation: 4)(20x+12)!=0

x∈R



We add all the numbers together, and all the variables

(+5/4)(20x+12)-(-35)=0

We add all the numbers together, and all the variables

(+5/4)(20x+12)+35=0

We multiply parentheses ..

(+100x^2+5/4*12)+35=0

We multiply all the terms by the denominator


(+100x^2+5+35*4*12)=0

We get rid of parentheses

100x^2+5+35*4*12=0

We add all the numbers together, and all the variables

100x^2+1685=0

a = 100; b = 0; c = +1685;
Δ = b2-4ac
Δ = 02-4·100·1685
Δ = -674000
Delta is less than zero, so there is no solution for the equation

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