Solve x*(3,4x)=x*3

$$x \cdot ( 3,4 x ) = x \cdot 3$$

Solve for x

$$x=\frac{15}{17}\approx 0,882352941$$

$$x=0$$

Steps Using Factoring

Multiply x and x to get x^{2}.

$$x^{2}\times 3,4=x\times 3$$

Subtract x\times 3 from both sides.

$$x^{2}\times 3,4-x\times 3=0$$

Multiply -1 and 3 to get -3.

$$x^{2}\times 3,4-3x=0$$

Factor out x.

$$x\left(3,4x-3\right)=0$$

To find equation solutions, solve x=0 and \frac{17x}{5}-3=0.

$$x=0 x=\frac{15}{17}$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$x^{2}\times 3,4=x\times 3$$

Subtract x\times 3 from both sides.

$$x^{2}\times 3,4-x\times 3=0$$

Multiply -1 and 3 to get -3.

$$x^{2}\times 3,4-3x=0$$

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.

$$3,4x^{2}-3x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3,4 for a, -3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-3\right)^{2}.

$$x=\frac{-\left(-3\right)±3}{2\times 3,4}$$

The opposite of -3 is 3.

$$x=\frac{3±3}{2\times 3,4}$$

Multiply 2 times 3,4.

$$x=\frac{3±3}{6,8}$$

Now solve the equation x=\frac{3±3}{6,8} when ± is plus. Add 3 to 3.

$$x=\frac{6}{6,8}$$

Divide 6 by 6,8 by multiplying 6 by the reciprocal of 6,8.

$$x=\frac{15}{17}$$

Now solve the equation x=\frac{3±3}{6,8} when ± is minus. Subtract 3 from 3.

$$x=\frac{0}{6,8}$$

Divide 0 by 6,8 by multiplying 0 by the reciprocal of 6,8.

$$x=0$$

The equation is now solved.

$$x=\frac{15}{17} x=0$$

Steps for Completing the Square

Multiply x and x to get x^{2}.

$$x^{2}\times 3,4=x\times 3$$

Subtract x\times 3 from both sides.

$$x^{2}\times 3,4-x\times 3=0$$

Multiply -1 and 3 to get -3.

$$x^{2}\times 3,4-3x=0$$

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.

$$3,4x^{2}-3x=0$$

Divide both sides of the equation by 3,4, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{3,4x^{2}-3x}{3,4}=\frac{0}{3,4}$$

Dividing by 3,4 undoes the multiplication by 3,4.

$$x^{2}+\left(-\frac{3}{3,4}\right)x=\frac{0}{3,4}$$

Divide -3 by 3,4 by multiplying -3 by the reciprocal of 3,4.

$$x^{2}-\frac{15}{17}x=\frac{0}{3,4}$$

Divide 0 by 3,4 by multiplying 0 by the reciprocal of 3,4.

$$x^{2}-\frac{15}{17}x=0$$

Divide -\frac{15}{17}, the coefficient of the x term, by 2 to get -\frac{15}{34}. Then add the square of -\frac{15}{34} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{15}{17}x+\left(-\frac{15}{34}\right)^{2}=\left(-\frac{15}{34}\right)^{2}$$

Square -\frac{15}{34} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{15}{17}x+\frac{225}{1156}=\frac{225}{1156}$$

Factor x^{2}-\frac{15}{17}x+\frac{225}{1156}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{15}{34}\right)^{2}=\frac{225}{1156}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{34}\right)^{2}}=\sqrt{\frac{225}{1156}}$$

Simplify.

$$x-\frac{15}{34}=\frac{15}{34} x-\frac{15}{34}=-\frac{15}{34}$$

Add \frac{15}{34} to both sides of the equation.

$$x=\frac{15}{17} x=0$$