Solve 2x*3+x/5=5

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

Solve for x

$$x = \frac{\sqrt{59} - 3}{2} \approx 2.340572874$$

$$x=\frac{-\sqrt{59}-3}{2}\approx -5.340572874$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 5.

$$2x\left(3+x\right)=25$$

Use the distributive property to multiply 2x by 3+x.

$$6x+2x^{2}=25$$

Subtract 25 from both sides.

$$6x+2x^{2}-25=0$$

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.

$$2x^{2}+6x-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 6 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 2\left(-25\right)}}{2\times 2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 2\left(-25\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-6±\sqrt{36-8\left(-25\right)}}{2\times 2}$$

Multiply -8 times -25.

$$x=\frac{-6±\sqrt{36+200}}{2\times 2}$$

Add 36 to 200.

$$x=\frac{-6±\sqrt{236}}{2\times 2}$$

Take the square root of 236.

$$x=\frac{-6±2\sqrt{59}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-6±2\sqrt{59}}{4}$$

Now solve the equation x=\frac{-6±2\sqrt{59}}{4} when ± is plus. Add -6 to 2\sqrt{59}.

$$x=\frac{2\sqrt{59}-6}{4}$$

Divide -6+2\sqrt{59} by 4.

$$x=\frac{\sqrt{59}-3}{2}$$

Now solve the equation x=\frac{-6±2\sqrt{59}}{4} when ± is minus. Subtract 2\sqrt{59} from -6.

$$x=\frac{-2\sqrt{59}-6}{4}$$

Divide -6-2\sqrt{59} by 4.

$$x=\frac{-\sqrt{59}-3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{59}-3}{2} x=\frac{-\sqrt{59}-3}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 5.

$$2x\left(3+x\right)=25$$

Use the distributive property to multiply 2x by 3+x.

$$6x+2x^{2}=25$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}+6x=25$$

Divide both sides by 2.

$$\frac{2x^{2}+6x}{2}=\frac{25}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{6}{2}x=\frac{25}{2}$$

Divide 6 by 2.

$$x^{2}+3x=\frac{25}{2}$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=\frac{25}{2}+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=\frac{25}{2}+\frac{9}{4}$$

Add \frac{25}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=\frac{59}{4}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{59}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{59}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{59}}{2} x+\frac{3}{2}=-\frac{\sqrt{59}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{59}-3}{2} x=\frac{-\sqrt{59}-3}{2}$$