3x/7+(x-3)X=10

Solution for 3x/7+(x-3)X=10 equation:

3x/7+(x-3)x=10

We move all terms to the left:

3x/7+(x-3)x-(10)=0

We multiply parentheses

x^2+3x/7-3x-10=0

We multiply all the terms by the denominator


x^2*7+3x-3x*7-10*7=0

We add all the numbers together, and all the variables

x^2*7+3x-3x*7-70=0

Wy multiply elements

7x^2+3x-21x-70=0

We add all the numbers together, and all the variables

7x^2-18x-70=0

a = 7; b = -18; c = -70;
Δ = b2-4ac
Δ = -182-4·7·(-70)
Δ = 2284
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:

$\sqrt{\Delta}=\sqrt{2284}=\sqrt{4*571}=\sqrt{4}*\sqrt{571}=2\sqrt{571}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{571}}{2*7}=\frac{18-2\sqrt{571}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{571}}{2*7}=\frac{18+2\sqrt{571}}{14}
$