Solve (x+5)(x+2)-3(x-4)^2>x(3-2x)+5

$$( x + 5 ) ( x + 2 ) - 3 ( x - 4 ) ^ { 2 } > x ( 3 - 2 x ) + 5$$

Solve for x

$$x>\frac{43}{28}$$

Solution Steps

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

$$x^{2}+7x+10-3\left(x-4\right)^{2}>x\left(3-2x\right)+5$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}+7x+10-3\left(x^{2}-8x+16\right)>x\left(3-2x\right)+5$$

Use the distributive property to multiply -3 by x^{2}-8x+16.

$$x^{2}+7x+10-3x^{2}+24x-48>x\left(3-2x\right)+5$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-2x^{2}+7x+10+24x-48>x\left(3-2x\right)+5$$

Combine 7x and 24x to get 31x.

$$-2x^{2}+31x+10-48>x\left(3-2x\right)+5$$

Subtract 48 from 10 to get -38.

$$-2x^{2}+31x-38>x\left(3-2x\right)+5$$

Use the distributive property to multiply x by 3-2x.

$$-2x^{2}+31x-38>3x-2x^{2}+5$$

Subtract 3x from both sides.

$$-2x^{2}+31x-38-3x>-2x^{2}+5$$

Combine 31x and -3x to get 28x.

$$-2x^{2}+28x-38>-2x^{2}+5$$

Add 2x^{2} to both sides.

$$-2x^{2}+28x-38+2x^{2}>5$$

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

$$28x-38>5$$

Add 38 to both sides.

$$28x>5+38$$

Add 5 and 38 to get 43.

$$28x>43$$

Divide both sides by 28. Since 28 is positive, the inequality direction remains the same.

$$x>\frac{43}{28}$$