Solve (-5/3)x^2-10x+6(-5/3)=5

$$( - \frac { 5 } { 3 } ) x ^ { 2 } - 10 x + 6 ( - \frac { 5 } { 3 } ) = 5$$

Solve for x

$$x=-3$$

Steps Using the Quadratic Formula

Multiply 6 and -\frac{5}{3} to get -10.

$$-\frac{5}{3}x^{2}-10x-10=5$$

Subtract 5 from both sides.

$$-\frac{5}{3}x^{2}-10x-10-5=0$$

Subtract 5 from -10 to get -15.

$$-\frac{5}{3}x^{2}-10x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{5}{3} for a, -10 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-\frac{5}{3}\right)\left(-15\right)}}{2\left(-\frac{5}{3}\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-\frac{5}{3}\right)\left(-15\right)}}{2\left(-\frac{5}{3}\right)}$$

Multiply -4 times -\frac{5}{3}.

$$x=\frac{-\left(-10\right)±\sqrt{100+\frac{20}{3}\left(-15\right)}}{2\left(-\frac{5}{3}\right)}$$

Multiply \frac{20}{3} times -15.

$$x=\frac{-\left(-10\right)±\sqrt{100-100}}{2\left(-\frac{5}{3}\right)}$$

Add 100 to -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2\left(-\frac{5}{3}\right)}$$

Take the square root of 0.

$$x=-\frac{-10}{2\left(-\frac{5}{3}\right)}$$

The opposite of -10 is 10.

$$x=\frac{10}{2\left(-\frac{5}{3}\right)}$$

Multiply 2 times -\frac{5}{3}.

$$x=\frac{10}{-\frac{10}{3}}$$

Divide 10 by -\frac{10}{3} by multiplying 10 by the reciprocal of -\frac{10}{3}.

$$x=-3$$

Steps for Completing the Square

Multiply 6 and -\frac{5}{3} to get -10.

$$-\frac{5}{3}x^{2}-10x-10=5$$

Add 10 to both sides.

$$-\frac{5}{3}x^{2}-10x=5+10$$

Add 5 and 10 to get 15.

$$-\frac{5}{3}x^{2}-10x=15$$

Divide both sides of the equation by -\frac{5}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{5}{3}x^{2}-10x}{-\frac{5}{3}}=\frac{15}{-\frac{5}{3}}$$

Dividing by -\frac{5}{3} undoes the multiplication by -\frac{5}{3}.

$$x^{2}+\left(-\frac{10}{-\frac{5}{3}}\right)x=\frac{15}{-\frac{5}{3}}$$

Divide -10 by -\frac{5}{3} by multiplying -10 by the reciprocal of -\frac{5}{3}.

$$x^{2}+6x=\frac{15}{-\frac{5}{3}}$$

Divide 15 by -\frac{5}{3} by multiplying 15 by the reciprocal of -\frac{5}{3}.

$$x^{2}+6x=-9$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-9+3^{2}$$

Square 3.

$$x^{2}+6x+9=-9+9$$

Add -9 to 9.

$$x^{2}+6x+9=0$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+3=0 x+3=0$$

Subtract 3 from both sides of the equation.

$$x=-3 x=-3$$

The equation is now solved. Solutions are the same.

$$x=-3$$