Solve (2x-5)^2+3(x-1)>4x(x-4)+15

$$( 2 x - 5 ) ^ { 2 } + 3 ( x - 1 ) > 4 x ( x - 4 ) + 15$$

Solve for x

$$x<7$$

Solution Steps

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

$$4x^{2}-20x+25+3\left(x-1\right)>4x\left(x-4\right)+15$$

Use the distributive property to multiply 3 by x-1.

$$4x^{2}-20x+25+3x-3>4x\left(x-4\right)+15$$

Combine -20x and 3x to get -17x.

$$4x^{2}-17x+25-3>4x\left(x-4\right)+15$$

Subtract 3 from 25 to get 22.

$$4x^{2}-17x+22>4x\left(x-4\right)+15$$

Use the distributive property to multiply 4x by x-4.

$$4x^{2}-17x+22>4x^{2}-16x+15$$

Subtract 4x^{2} from both sides.

$$4x^{2}-17x+22-4x^{2}>-16x+15$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-17x+22>-16x+15$$

Add 16x to both sides.

$$-17x+22+16x>15$$

Combine -17x and 16x to get -x.

$$-x+22>15$$

Subtract 22 from both sides.

$$-x>15-22$$

Subtract 22 from 15 to get -7.

$$-x>-7$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x<\frac{-7}{-1}$$

Fraction \frac{-7}{-1} can be simplified to 7 by removing the negative sign from both the numerator and the denominator.

$$x<7$$