(3x-1)+(8x-26)x=

Solution for (3x-1)+(8x-26)x= equation:

(3x-1)+(8x-26)x=

We move all terms to the left:

(3x-1)+(8x-26)x-()=0

We add all the numbers together, and all the variables

(3x-1)+(8x-26)x=0

We multiply parentheses

8x^2+(3x-1)-26x=0

We get rid of parentheses

8x^2+3x-26x-1=0

We add all the numbers together, and all the variables

8x^2-23x-1=0

a = 8; b = -23; c = -1;
Δ = b2-4ac
Δ = -232-4·8·(-1)
Δ = 561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{561}}{2*8}=\frac{23-\sqrt{561}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{561}}{2*8}=\frac{23+\sqrt{561}}{16}
$