49x^2-140x+100=(7x-10)

Solution for 49x^2-140x+100=(7x-10) equation:

49x^2-140x+100=(7x-10)

We move all terms to the left:

49x^2-140x+100-((7x-10))=0


We calculate terms in parentheses: -((7x-10)), so:

(7x-10)

We get rid of parentheses

7x-10

Back to the equation:

-(7x-10)


We get rid of parentheses

49x^2-140x-7x+10+100=0

We add all the numbers together, and all the variables

49x^2-147x+110=0

a = 49; b = -147; c = +110;
Δ = b2-4ac
Δ = -1472-4·49·110
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-147)-7}{2*49}=\frac{140}{98}
=1+3/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-147)+7}{2*49}=\frac{154}{98}
=1+4/7
$