(1/2)(10y+4)=17

Solution for (1/2)(10y+4)=17 equation:

(1/2)(10y+4)=17

We move all terms to the left:

(1/2)(10y+4)-(17)=0


Domain of the equation: 2)(10y+4)!=0

y∈R


We add all the numbers together, and all the variables

(+1/2)(10y+4)-17=0

We multiply parentheses ..

(+10y^2+1/2*4)-17=0

We multiply all the terms by the denominator


(+10y^2+1-17*2*4)=0

We get rid of parentheses

10y^2+1-17*2*4=0

We add all the numbers together, and all the variables

10y^2-135=0

a = 10; b = 0; c = -135;
Δ = b2-4ac
Δ = 02-4·10·(-135)
Δ = 5400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5400}=\sqrt{900*6}=\sqrt{900}*\sqrt{6}=30\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30\sqrt{6}}{2*10}=\frac{0-30\sqrt{6}}{20}
=-\frac{30\sqrt{6}}{20}
=-\frac{3\sqrt{6}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30\sqrt{6}}{2*10}=\frac{0+30\sqrt{6}}{20}
=\frac{30\sqrt{6}}{20}
=\frac{3\sqrt{6}}{2}
$