Solve (y-5)(y+2)-(y-7)^2>18

$$( y - 5 ) ( y + 2 ) - ( y - 7 ) ^ { 2 } > 18$$

Solve for y

$$y>7$$

Solution Steps

Use the distributive property to multiply y-5 by y+2 and combine like terms.

$$y^{2}-3y-10-\left(y-7\right)^{2}>18$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-7\right)^{2}.

$$y^{2}-3y-10-\left(y^{2}-14y+49\right)>18$$

To find the opposite of y^{2}-14y+49, find the opposite of each term.

$$y^{2}-3y-10-y^{2}+14y-49>18$$

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

$$-3y-10+14y-49>18$$

Combine -3y and 14y to get 11y.

$$11y-10-49>18$$

Subtract 49 from -10 to get -59.

$$11y-59>18$$

Add 59 to both sides.

$$11y>18+59$$

Add 18 and 59 to get 77.

$$11y>77$$

Divide both sides by 11. Since 11 is positive, the inequality direction remains the same.

$$y>\frac{77}{11}$$

Divide 77 by 11 to get 7.

$$y>7$$