Solve (x-5)(x+2)-(x+3)^2>7-14x

$$( x - 5 ) ( x + 2 ) - ( x + 3 ) ^ { 2 } > 7 - 14 x$$

Solve for x

$$x>\frac{26}{5}$$

Solution Steps

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

$$x^{2}-3x-10-\left(x+3\right)^{2}>7-14x$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}-3x-10-\left(x^{2}+6x+9\right)>7-14x$$

To find the opposite of x^{2}+6x+9, find the opposite of each term.

$$x^{2}-3x-10-x^{2}-6x-9>7-14x$$

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

$$-3x-10-6x-9>7-14x$$

Combine -3x and -6x to get -9x.

$$-9x-10-9>7-14x$$

Subtract 9 from -10 to get -19.

$$-9x-19>7-14x$$

Add 14x to both sides.

$$-9x-19+14x>7$$

Combine -9x and 14x to get 5x.

$$5x-19>7$$

Add 19 to both sides.

$$5x>7+19$$

Add 7 and 19 to get 26.

$$5x>26$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

$$x>\frac{26}{5}$$