12+(1/5)(10x+5)=59

Solution for 12+(1/5)(10x+5)=59 equation:

12+(1/5)(10x+5)=59

We move all terms to the left:

12+(1/5)(10x+5)-(59)=0


Domain of the equation: 5)(10x+5)!=0

x∈R


We add all the numbers together, and all the variables

(+1/5)(10x+5)+12-59=0

We add all the numbers together, and all the variables

(+1/5)(10x+5)-47=0

We multiply parentheses ..

(+10x^2+1/5*5)-47=0

We multiply all the terms by the denominator


(+10x^2+1-47*5*5)=0

We get rid of parentheses

10x^2+1-47*5*5=0

We add all the numbers together, and all the variables

10x^2-1174=0

a = 10; b = 0; c = -1174;
Δ = b2-4ac
Δ = 02-4·10·(-1174)
Δ = 46960
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:

$\sqrt{\Delta}=\sqrt{46960}=\sqrt{16*2935}=\sqrt{16}*\sqrt{2935}=4\sqrt{2935}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2935}}{2*10}=\frac{0-4\sqrt{2935}}{20}
=-\frac{4\sqrt{2935}}{20}
=-\frac{\sqrt{2935}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2935}}{2*10}=\frac{0+4\sqrt{2935}}{20}
=\frac{4\sqrt{2935}}{20}
=\frac{\sqrt{2935}}{5}
$