(7w^2-2w+1)=(-6w^2+3w+5)

Solution for (7w^2-2w+1)=(-6w^2+3w+5) equation:

(7w^2-2w+1)=(-6w^2+3w+5)

We move all terms to the left:

(7w^2-2w+1)-((-6w^2+3w+5))=0

We get rid of parentheses

-((-6w^2+3w+5))+7w^2-2w+1=0


We calculate terms in parentheses: -((-6w^2+3w+5)), so:

(-6w^2+3w+5)

We get rid of parentheses

-6w^2+3w+5

Back to the equation:

-(-6w^2+3w+5)


We add all the numbers together, and all the variables

7w^2-(-6w^2+3w+5)-2w+1=0

We get rid of parentheses

7w^2+6w^2-3w-2w-5+1=0

We add all the numbers together, and all the variables

13w^2-5w-4=0

a = 13; b = -5; c = -4;
Δ = b2-4ac
Δ = -52-4·13·(-4)
Δ = 233
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{233}}{2*13}=\frac{5-\sqrt{233}}{26}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{233}}{2*13}=\frac{5+\sqrt{233}}{26}
$