Solve sqrt{3x^2-5x+6}=2(x+2)

$$\sqrt { 3 x ^ { 2 } - 5 x + 6 } = 2 ( x + 2 )$$

Solve for x

$$x=\frac{\sqrt{401}-21}{2}\approx -0.487507803$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x^{2}-5x+6}\right)^{2}=\left(2\left(x+2\right)\right)^{2}$$

Calculate \sqrt{3x^{2}-5x+6} to the power of 2 and get 3x^{2}-5x+6.

$$3x^{2}-5x+6=\left(2\left(x+2\right)\right)^{2}$$

Use the distributive property to multiply 2 by x+2.

$$3x^{2}-5x+6=\left(2x+4\right)^{2}$$

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

$$3x^{2}-5x+6=4x^{2}+16x+16$$

Subtract 4x^{2} from both sides.

$$3x^{2}-5x+6-4x^{2}=16x+16$$

Combine 3x^{2} and -4x^{2} to get -x^{2}.

$$-x^{2}-5x+6=16x+16$$

Subtract 16x from both sides.

$$-x^{2}-5x+6-16x=16$$

Combine -5x and -16x to get -21x.

$$-x^{2}-21x+6=16$$

Subtract 16 from both sides.

$$-x^{2}-21x+6-16=0$$

Subtract 16 from 6 to get -10.

$$-x^{2}-21x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -21 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\left(-1\right)\left(-10\right)}}{2\left(-1\right)}$$

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\left(-1\right)\left(-10\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-21\right)±\sqrt{441+4\left(-10\right)}}{2\left(-1\right)}$$

Multiply 4 times -10.

$$x=\frac{-\left(-21\right)±\sqrt{441-40}}{2\left(-1\right)}$$

Add 441 to -40.

$$x=\frac{-\left(-21\right)±\sqrt{401}}{2\left(-1\right)}$$

The opposite of -21 is 21.

$$x=\frac{21±\sqrt{401}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{21±\sqrt{401}}{-2}$$

Now solve the equation x=\frac{21±\sqrt{401}}{-2} when ± is plus. Add 21 to \sqrt{401}.

$$x=\frac{\sqrt{401}+21}{-2}$$

Divide 21+\sqrt{401} by -2.

$$x=\frac{-\sqrt{401}-21}{2}$$

Now solve the equation x=\frac{21±\sqrt{401}}{-2} when ± is minus. Subtract \sqrt{401} from 21.

$$x=\frac{21-\sqrt{401}}{-2}$$

Divide 21-\sqrt{401} by -2.

$$x=\frac{\sqrt{401}-21}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{401}-21}{2} x=\frac{\sqrt{401}-21}{2}$$

Substitute \frac{-\sqrt{401}-21}{2} for x in the equation \sqrt{3x^{2}-5x+6}=2\left(x+2\right).

$$\sqrt{3\times \left(\frac{-\sqrt{401}-21}{2}\right)^{2}-5\times \frac{-\sqrt{401}-21}{2}+6}=2\left(\frac{-\sqrt{401}-21}{2}+2\right)$$

Simplify. The value x=\frac{-\sqrt{401}-21}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$401^{\frac{1}{2}}+17=-401^{\frac{1}{2}}-17$$

Substitute \frac{\sqrt{401}-21}{2} for x in the equation \sqrt{3x^{2}-5x+6}=2\left(x+2\right).

$$\sqrt{3\times \left(\frac{\sqrt{401}-21}{2}\right)^{2}-5\times \frac{\sqrt{401}-21}{2}+6}=2\left(\frac{\sqrt{401}-21}{2}+2\right)$$

Simplify. The value x=\frac{\sqrt{401}-21}{2} satisfies the equation.

$$401^{\frac{1}{2}}-17=401^{\frac{1}{2}}-17$$

Equation \sqrt{3x^{2}-5x+6}=2\left(x+2\right) has a unique solution.

$$x=\frac{\sqrt{401}-21}{2}$$