(x-3/2)(2x+5)=25

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Solution for (x-3/2)(2x+5)=25 equation: 



(x-3/2)(2x+5)=25

We move all terms to the left:

(x-3/2)(2x+5)-(25)=0



Domain of the equation: 2)(2x+5)!=0

x∈R



We add all the numbers together, and all the variables

(+x-3/2)(2x+5)-25=0

We multiply parentheses ..

(+2x^2+5x-6x^2-3/2*5)-25=0

We multiply all the terms by the denominator


(+2x^2+5x-6x^2-3-25*2*5)=0

We get rid of parentheses

2x^2-6x^2+5x-3-25*2*5=0

We add all the numbers together, and all the variables

-4x^2+5x-253=0

a = -4; b = 5; c = -253;
Δ = b2-4ac
Δ = 52-4·(-4)·(-253)
Δ = -4023
Delta is less than zero, so there is no solution for the equation

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