Solve 0=-4*9x^2+10*4x-3

$$0 = - 4 \cdot 9 x ^ { 2 } + 10 \cdot 4 x - 3$$

Solve for x

$$x=\frac{\sqrt{73}}{18}+\frac{5}{9}\approx 1.03022243$$

$$x=-\frac{\sqrt{73}}{18}+\frac{5}{9}\approx 0.080888681$$

Steps Using the Quadratic Formula

Do the multiplications.

$$0=-36x^{2}+40x-3$$

Swap sides so that all variable terms are on the left hand side.

$$-36x^{2}+40x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -36 for a, 40 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\left(-36\right)\left(-3\right)}}{2\left(-36\right)}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\left(-36\right)\left(-3\right)}}{2\left(-36\right)}$$

Multiply -4 times -36.

$$x=\frac{-40±\sqrt{1600+144\left(-3\right)}}{2\left(-36\right)}$$

Multiply 144 times -3.

$$x=\frac{-40±\sqrt{1600-432}}{2\left(-36\right)}$$

Add 1600 to -432.

$$x=\frac{-40±\sqrt{1168}}{2\left(-36\right)}$$

Take the square root of 1168.

$$x=\frac{-40±4\sqrt{73}}{2\left(-36\right)}$$

Multiply 2 times -36.

$$x=\frac{-40±4\sqrt{73}}{-72}$$

Now solve the equation x=\frac{-40±4\sqrt{73}}{-72} when ± is plus. Add -40 to 4\sqrt{73}.

$$x=\frac{4\sqrt{73}-40}{-72}$$

Divide -40+4\sqrt{73} by -72.

$$x=-\frac{\sqrt{73}}{18}+\frac{5}{9}$$

Now solve the equation x=\frac{-40±4\sqrt{73}}{-72} when ± is minus. Subtract 4\sqrt{73} from -40.

$$x=\frac{-4\sqrt{73}-40}{-72}$$

Divide -40-4\sqrt{73} by -72.

$$x=\frac{\sqrt{73}}{18}+\frac{5}{9}$$

The equation is now solved.

$$x=-\frac{\sqrt{73}}{18}+\frac{5}{9} x=\frac{\sqrt{73}}{18}+\frac{5}{9}$$

Steps for Completing the Square

Do the multiplications.

$$0=-36x^{2}+40x-3$$

Swap sides so that all variable terms are on the left hand side.

$$-36x^{2}+40x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$-36x^{2}+40x=3$$

Divide both sides by -36.

$$\frac{-36x^{2}+40x}{-36}=\frac{3}{-36}$$

Dividing by -36 undoes the multiplication by -36.

$$x^{2}+\frac{40}{-36}x=\frac{3}{-36}$$

Reduce the fraction \frac{40}{-36} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{10}{9}x=\frac{3}{-36}$$

Reduce the fraction \frac{3}{-36} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{10}{9}x=-\frac{1}{12}$$

Divide -\frac{10}{9}, the coefficient of the x term, by 2 to get -\frac{5}{9}. Then add the square of -\frac{5}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{9}x+\left(-\frac{5}{9}\right)^{2}=-\frac{1}{12}+\left(-\frac{5}{9}\right)^{2}$$

Square -\frac{5}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=-\frac{1}{12}+\frac{25}{81}$$

Add -\frac{1}{12} to \frac{25}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=\frac{73}{324}$$

Factor x^{2}-\frac{10}{9}x+\frac{25}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{9}\right)^{2}=\frac{73}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{9}\right)^{2}}=\sqrt{\frac{73}{324}}$$

Simplify.

$$x-\frac{5}{9}=\frac{\sqrt{73}}{18} x-\frac{5}{9}=-\frac{\sqrt{73}}{18}$$

Add \frac{5}{9} to both sides of the equation.

$$x=\frac{\sqrt{73}}{18}+\frac{5}{9} x=-\frac{\sqrt{73}}{18}+\frac{5}{9}$$