4(1/x-4)^2-(3/4-x)-10=0

Solution for 4(1/x-4)^2-(3/4-x)-10=0 equation:

4(1/x-4)^2-(3/4-x)-10=0


Domain of the equation: x-4)^2!=0

x∈R



Domain of the equation: 4-x)!=0

We move all terms containing x to the left, all other terms to the right

-x)!=-4

x!=-4/1

x!=-4

x∈R


We add all the numbers together, and all the variables

4(1/x-4)^2-(-1x+3/4)-10=0

We get rid of parentheses

4(1/x-4)^2+1x-10-3/4=0

We calculate fractions


1x+()/x+(-3x)/x-10=0

We add all the numbers together, and all the variables

x+()/x+(-3x)/x-10=0

We multiply all the terms by the denominator


x*x+(-3x)-10*x+()=0

We add all the numbers together, and all the variables

-10x+x*x+(-3x)=0

Wy multiply elements

x^2-10x+(-3x)=0

We get rid of parentheses

x^2-10x-3x=0

We add all the numbers together, and all the variables

x^2-13x=0

a = 1; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·1·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*1}=\frac{26}{2}
=13
$