Solve 114=-38x^2+152x

$$114=-38 { x }^{ 2 } +152x$$

Solve for x

$$x=1$$

$$x=3$$

Steps Using the Quadratic Formula

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

$$-38x^{2}+152x=114$$

Subtract 114 from both sides.

$$-38x^{2}+152x-114=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -38 for a, 152 for b, and -114 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-152±\sqrt{152^{2}-4\left(-38\right)\left(-114\right)}}{2\left(-38\right)}$$

Square 152.

$$x=\frac{-152±\sqrt{23104-4\left(-38\right)\left(-114\right)}}{2\left(-38\right)}$$

Multiply -4 times -38.

$$x=\frac{-152±\sqrt{23104+152\left(-114\right)}}{2\left(-38\right)}$$

Multiply 152 times -114.

$$x=\frac{-152±\sqrt{23104-17328}}{2\left(-38\right)}$$

Add 23104 to -17328.

$$x=\frac{-152±\sqrt{5776}}{2\left(-38\right)}$$

Take the square root of 5776.

$$x=\frac{-152±76}{2\left(-38\right)}$$

Multiply 2 times -38.

$$x=\frac{-152±76}{-76}$$

Now solve the equation x=\frac{-152±76}{-76} when ± is plus. Add -152 to 76.

$$x=-\frac{76}{-76}$$

Divide -76 by -76.

$$x=1$$

Now solve the equation x=\frac{-152±76}{-76} when ± is minus. Subtract 76 from -152.

$$x=-\frac{228}{-76}$$

Divide -228 by -76.

$$x=3$$

The equation is now solved.

$$x=1 x=3$$

Steps for Completing the Square

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

$$-38x^{2}+152x=114$$

Divide both sides by -38.

$$\frac{-38x^{2}+152x}{-38}=\frac{114}{-38}$$

Dividing by -38 undoes the multiplication by -38.

$$x^{2}+\frac{152}{-38}x=\frac{114}{-38}$$

Divide 152 by -38.

$$x^{2}-4x=\frac{114}{-38}$$

Divide 114 by -38.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

$$x^{2}-4x+4=-3+4$$

Add -3 to 4.

$$x^{2}-4x+4=1$$

Factor x^{2}-4x+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}.

$$\left(x-2\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$