(y-3)(4y=7)

Solution for (y-3)(4y=7) equation:

(y-3)(4y=7)

We move all terms to the left:

(y-3)(4y-(7))=0

We multiply parentheses ..

(+4y^2-7y-12y+21)=0

We get rid of parentheses

4y^2-7y-12y+21=0

We add all the numbers together, and all the variables

4y^2-19y+21=0

a = 4; b = -19; c = +21;
Δ = b2-4ac
Δ = -192-4·4·21
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5}{2*4}=\frac{14}{8}
=1+3/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5}{2*4}=\frac{24}{8}
=3
$