Solve 5x^2=3left(0.40+xright)^2

$$5 { x }^{ 2 } =3 { \left(0.40+x \right) }^{ 2 }$$

Solve for x

$$x = \frac{\sqrt{15} + 3}{5} \approx 1.374596669$$

$$x=\frac{3-\sqrt{15}}{5}\approx -0.174596669$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(0.4+x\right)^{2}.

$$5x^{2}=3\left(0.16+0.8x+x^{2}\right)$$

Use the distributive property to multiply 3 by 0.16+0.8x+x^{2}.

$$5x^{2}=0.48+2.4x+3x^{2}$$

Subtract 0.48 from both sides.

$$5x^{2}-0.48=2.4x+3x^{2}$$

Subtract 2.4x from both sides.

$$5x^{2}-0.48-2.4x=3x^{2}$$

Subtract 3x^{2} from both sides.

$$5x^{2}-0.48-2.4x-3x^{2}=0$$

Combine 5x^{2} and -3x^{2} to get 2x^{2}.

$$2x^{2}-0.48-2.4x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-2.4x-0.48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -2.4 for b, and -0.48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2.4\right)±\sqrt{\left(-2.4\right)^{2}-4\times 2\left(-0.48\right)}}{2\times 2}$$

Square -2.4 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-2.4\right)±\sqrt{5.76-4\times 2\left(-0.48\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-2.4\right)±\sqrt{5.76-8\left(-0.48\right)}}{2\times 2}$$

Multiply -8 times -0.48.

$$x=\frac{-\left(-2.4\right)±\sqrt{\frac{144+96}{25}}}{2\times 2}$$

Add 5.76 to 3.84 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-2.4\right)±\sqrt{9.6}}{2\times 2}$$

Take the square root of 9.6.

$$x=\frac{-\left(-2.4\right)±\frac{4\sqrt{15}}{5}}{2\times 2}$$

The opposite of -2.4 is 2.4.

$$x=\frac{2.4±\frac{4\sqrt{15}}{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{2.4±\frac{4\sqrt{15}}{5}}{4}$$

Now solve the equation x=\frac{2.4±\frac{4\sqrt{15}}{5}}{4} when ± is plus. Add 2.4 to \frac{4\sqrt{15}}{5}.

$$x=\frac{4\sqrt{15}+12}{4\times 5}$$

Divide \frac{12+4\sqrt{15}}{5} by 4.

$$x=\frac{\sqrt{15}+3}{5}$$

Now solve the equation x=\frac{2.4±\frac{4\sqrt{15}}{5}}{4} when ± is minus. Subtract \frac{4\sqrt{15}}{5} from 2.4.

$$x=\frac{12-4\sqrt{15}}{4\times 5}$$

Divide \frac{12-4\sqrt{15}}{5} by 4.

$$x=\frac{3-\sqrt{15}}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{15}+3}{5} x=\frac{3-\sqrt{15}}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(0.4+x\right)^{2}.

$$5x^{2}=3\left(0.16+0.8x+x^{2}\right)$$

Use the distributive property to multiply 3 by 0.16+0.8x+x^{2}.

$$5x^{2}=0.48+2.4x+3x^{2}$$

Subtract 2.4x from both sides.

$$5x^{2}-2.4x=0.48+3x^{2}$$

Subtract 3x^{2} from both sides.

$$5x^{2}-2.4x-3x^{2}=0.48$$

Combine 5x^{2} and -3x^{2} to get 2x^{2}.

$$2x^{2}-2.4x=0.48$$

Divide both sides by 2.

$$\frac{2x^{2}-2.4x}{2}=\frac{0.48}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{2.4}{2}\right)x=\frac{0.48}{2}$$

Divide -2.4 by 2.

$$x^{2}-1.2x=\frac{0.48}{2}$$

Divide 0.48 by 2.

$$x^{2}-1.2x=0.24$$

Divide -1.2, the coefficient of the x term, by 2 to get -0.6. Then add the square of -0.6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-1.2x+\left(-0.6\right)^{2}=0.24+\left(-0.6\right)^{2}$$

Square -0.6 by squaring both the numerator and the denominator of the fraction.

$$x^{2}-1.2x+0.36=\frac{6+9}{25}$$

Add 0.24 to 0.36 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-1.2x+0.36=0.6$$

Factor x^{2}-1.2x+0.36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-0.6\right)^{2}=0.6$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-0.6\right)^{2}}=\sqrt{0.6}$$

Simplify.

$$x-0.6=\frac{\sqrt{15}}{5} x-0.6=-\frac{\sqrt{15}}{5}$$

Add 0.6 to both sides of the equation.

$$x=\frac{\sqrt{15}+3}{5} x=\frac{3-\sqrt{15}}{5}$$