What is the answer to 4y^2+7y=10+(2y-3)^2 ?

The answer to 4y^2+7y=10+(2y-3)^2 is y=1

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4y^2+7y=10+(2y-3)^2

4 y^{2} + 7 y = 10 + (2 y - 3)^{2}

y = 1

Solution steps

4 y^{2} + 7 y = 10 + (2 y - 3)^{2}

Expand

10 + (2 y - 3)^{2} : 4 y^{2} - 12 y + 19

4 y^{2} + 7 y = 4 y^{2} - 12 y + 19

Subtract

4 y^{2} - 12 y

from both sides

4 y^{2} + 7 y - (4 y^{2} - 12 y) = 4 y^{2} - 12 y + 19 - (4 y^{2} - 12 y)

Simplify

19 y = 19

Divide both sides by

19

y = 1

y, x - y = 7

\frac{( 5 x + 6 )}{2} = 3 - (4 x + 12)

ln (2) = 2 x + 3

- \frac{1}{2} d = 3 (\frac{2}{3} d - 1) + 2 (\frac{1}{6} d + 2)

\frac{4}{9} x + \frac{1}{5} x = 87

What is the answer to 4y^2+7y=10+(2y-3)^2 ?
The answer to 4y^2+7y=10+(2y-3)^2 is y=1
Solve for y: 4y^2+7y=10+(2y-3)^2
The solution is y=1