Solve (2x-3)^2+sqrt{25}=50

$$( 2 x - 3 ) ^ { 2 } + \sqrt { 25 } = 50$$

Solve for x

$$x = \frac{3 \sqrt{5} + 3}{2} \approx 4.854101966$$

$$x=\frac{3-3\sqrt{5}}{2}\approx -1.854101966$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9+\sqrt{25}=50$$

Calculate the square root of 25 and get 5.

$$4x^{2}-12x+9+5=50$$

Add 9 and 5 to get 14.

$$4x^{2}-12x+14=50$$

Subtract 50 from both sides.

$$4x^{2}-12x+14-50=0$$

Subtract 50 from 14 to get -36.

$$4x^{2}-12x-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\left(-36\right)}}{2\times 4}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\left(-36\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\left(-36\right)}}{2\times 4}$$

Multiply -16 times -36.

$$x=\frac{-\left(-12\right)±\sqrt{144+576}}{2\times 4}$$

Add 144 to 576.

$$x=\frac{-\left(-12\right)±\sqrt{720}}{2\times 4}$$

Take the square root of 720.

$$x=\frac{-\left(-12\right)±12\sqrt{5}}{2\times 4}$$

The opposite of -12 is 12.

$$x=\frac{12±12\sqrt{5}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{12±12\sqrt{5}}{8}$$

Now solve the equation x=\frac{12±12\sqrt{5}}{8} when ± is plus. Add 12 to 12\sqrt{5}.

$$x=\frac{12\sqrt{5}+12}{8}$$

Divide 12+12\sqrt{5} by 8.

$$x=\frac{3\sqrt{5}+3}{2}$$

Now solve the equation x=\frac{12±12\sqrt{5}}{8} when ± is minus. Subtract 12\sqrt{5} from 12.

$$x=\frac{12-12\sqrt{5}}{8}$$

Divide 12-12\sqrt{5} by 8.

$$x=\frac{3-3\sqrt{5}}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{5}+3}{2} x=\frac{3-3\sqrt{5}}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9+\sqrt{25}=50$$

Calculate the square root of 25 and get 5.

$$4x^{2}-12x+9+5=50$$

Add 9 and 5 to get 14.

$$4x^{2}-12x+14=50$$

Subtract 14 from both sides.

$$4x^{2}-12x=50-14$$

Subtract 14 from 50 to get 36.

$$4x^{2}-12x=36$$

Divide both sides by 4.

$$\frac{4x^{2}-12x}{4}=\frac{36}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{12}{4}\right)x=\frac{36}{4}$$

Divide -12 by 4.

$$x^{2}-3x=\frac{36}{4}$$

Divide 36 by 4.

$$x^{2}-3x=9$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=9+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=9+\frac{9}{4}$$

Add 9 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{45}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{45}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{45}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{3\sqrt{5}}{2} x-\frac{3}{2}=-\frac{3\sqrt{5}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3\sqrt{5}+3}{2} x=\frac{3-3\sqrt{5}}{2}$$