Solve 3x^2+6geq(3x+1)(x-5)

$$3 x ^ { 2 } + 6 \geq ( 3 x + 1 ) ( x - 5 )$$

Solve for x

$$x\geq -\frac{11}{14}$$

Solution Steps

Use the distributive property to multiply 3x+1 by x-5 and combine like terms.

$$3x^{2}+6\geq 3x^{2}-14x-5$$

Subtract 3x^{2} from both sides.

$$3x^{2}+6-3x^{2}\geq -14x-5$$

Combine 3x^{2} and -3x^{2} to get 0.

$$6\geq -14x-5$$

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$-14x-5\leq 6$$

Add 5 to both sides.

$$-14x\leq 6+5$$

Add 6 and 5 to get 11.

$$-14x\leq 11$$

Divide both sides by -14. Since -14 is negative, the inequality direction is changed.

$$x\geq -\frac{11}{14}$$