Solve x^2+6x+18<x^2-10x+26

$${ x }^{ 2 } +6x+18 < { x }^{ 2 } -10x+26$$

Solve for x

$$x<\frac{1}{2}$$

Solution Steps

Subtract x^{2} from both sides.

$$x^{2}+6x+18-x^{2}<-10x+26$$

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

$$6x+18<-10x+26$$

Add 10x to both sides.

$$6x+18+10x<26$$

Combine 6x and 10x to get 16x.

$$16x+18<26$$

Subtract 18 from both sides.

$$16x<26-18$$

Subtract 18 from 26 to get 8.

$$16x<8$$

Divide both sides by 16. Since 16 is positive, the inequality direction remains the same.

$$x<\frac{8}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$x<\frac{1}{2}$$