(10-3y)(10-3y)/4+y*y=17

Solution for (10-3y)(10-3y)/4+y*y=17 equation:

(10-3y)(10-3y)/4+y*y=17

We move all terms to the left:

(10-3y)(10-3y)/4+y*y-(17)=0

We add all the numbers together, and all the variables

(-3y+10)(-3y+10)/4+y*y-17=0

Wy multiply elements

y^2+(-3y+10)(-3y+10)/4-17=0

We multiply parentheses ..

y^2+(+9y^2-30y-30y+100)/4-17=0

We multiply all the terms by the denominator


(+9y^2-30y-30y+100)+y^2*4-17*4=0

We add all the numbers together, and all the variables

(+9y^2-30y-30y+100)+y^2*4-68=0

Wy multiply elements

(+9y^2-30y-30y+100)+4y^2-68=0

We get rid of parentheses

9y^2+4y^2-30y-30y+100-68=0

We add all the numbers together, and all the variables

13y^2-60y+32=0

a = 13; b = -60; c = +32;
Δ = b2-4ac
Δ = -602-4·13·32
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-44}{2*13}=\frac{16}{26}
=8/13
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+44}{2*13}=\frac{104}{26}
=4
$