Solve 8y+left(2-5yright)^2+7yleft(-5y+2right)+10y^2+(2-5y)=3

$$8y+ { \left(2-5y \right) }^{ 2 } +7y \left( -5y+2 \right) +10 { y }^{ 2 } +(2-5y) = 3$$

Solve for y

$$y=1$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-5y\right)^{2}.

$$8y+4-20y+25y^{2}+7y\left(-5y+2\right)+10y^{2}+2-5y=3$$

Combine 8y and -20y to get -12y.

$$-12y+4+25y^{2}+7y\left(-5y+2\right)+10y^{2}+2-5y=3$$

Use the distributive property to multiply 7y by -5y+2.

$$-12y+4+25y^{2}-35y^{2}+14y+10y^{2}+2-5y=3$$

Combine 25y^{2} and -35y^{2} to get -10y^{2}.

$$-12y+4-10y^{2}+14y+10y^{2}+2-5y=3$$

Combine -12y and 14y to get 2y.

$$2y+4-10y^{2}+10y^{2}+2-5y=3$$

Combine -10y^{2} and 10y^{2} to get 0.

$$2y+4+2-5y=3$$

Add 4 and 2 to get 6.

$$2y+6-5y=3$$

Combine 2y and -5y to get -3y.

$$-3y+6=3$$

Subtract 6 from both sides.

$$-3y=3-6$$

Subtract 6 from 3 to get -3.

$$-3y=-3$$

Divide both sides by -3.

$$y=\frac{-3}{-3}$$

Divide -3 by -3 to get 1.

$$y=1$$