-(1/2x-3)+4(x+5)=0

Solution for -(1/2x-3)+4(x+5)=0 equation:

-(1/2x-3)+4(x+5)=0


Domain of the equation: 2x-3)!=0

x∈R


We multiply parentheses

-(1/2x-3)+4x+20=0

We get rid of parentheses

-1/2x+4x+3+20=0

We multiply all the terms by the denominator


4x*2x+3*2x+20*2x-1=0

Wy multiply elements

8x^2+6x+40x-1=0

We add all the numbers together, and all the variables

8x^2+46x-1=0

a = 8; b = 46; c = -1;
Δ = b2-4ac
Δ = 462-4·8·(-1)
Δ = 2148
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2148}=\sqrt{4*537}=\sqrt{4}*\sqrt{537}=2\sqrt{537}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-2\sqrt{537}}{2*8}=\frac{-46-2\sqrt{537}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+2\sqrt{537}}{2*8}=\frac{-46+2\sqrt{537}}{16}
$