Solve -3x^2+2x-1=-97

$$-3 { x }^{ 2 } +2x-1=-97$$

Solve for x

$$x = -\frac{16}{3} = -5\frac{1}{3} \approx -5.333333333$$

$$x=6$$

Steps Using Factoring By Grouping

Add 97 to both sides.

$$-3x^{2}+2x-1+97=0$$

Add -1 and 97 to get 96.

$$-3x^{2}+2x+96=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+96. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3\times 96=-288$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -288.

$$-1,288 -2,144 -3,96 -4,72 -6,48 -8,36 -9,32 -12,24 -16,18$$

Calculate the sum for each pair.

$$-1+288=287 -2+144=142 -3+96=93 -4+72=68 -6+48=42 -8+36=28 -9+32=23 -12+24=12 -16+18=2$$

The solution is the pair that gives sum 2.

$$a=18 b=-16$$

Rewrite -3x^{2}+2x+96 as \left(-3x^{2}+18x\right)+\left(-16x+96\right).

$$\left(-3x^{2}+18x\right)+\left(-16x+96\right)$$

Factor out 3x in the first and 16 in the second group.

$$3x\left(-x+6\right)+16\left(-x+6\right)$$

Factor out common term -x+6 by using distributive property.

$$\left(-x+6\right)\left(3x+16\right)$$

To find equation solutions, solve -x+6=0 and 3x+16=0.

$$x=6 x=-\frac{16}{3}$$

Steps Using the Quadratic Formula

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.

$$-3x^{2}+2x-1=-97$$

Add 97 to both sides of the equation.

$$-3x^{2}+2x-1-\left(-97\right)=-97-\left(-97\right)$$

Subtracting -97 from itself leaves 0.

$$-3x^{2}+2x-1-\left(-97\right)=0$$

Subtract -97 from -1.

$$-3x^{2}+2x+96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 2 for b, and 96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-3\right)\times 96}}{2\left(-3\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-3\right)\times 96}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-2±\sqrt{4+12\times 96}}{2\left(-3\right)}$$

Multiply 12 times 96.

$$x=\frac{-2±\sqrt{4+1152}}{2\left(-3\right)}$$

Add 4 to 1152.

$$x=\frac{-2±\sqrt{1156}}{2\left(-3\right)}$$

Take the square root of 1156.

$$x=\frac{-2±34}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-2±34}{-6}$$

Now solve the equation x=\frac{-2±34}{-6} when ± is plus. Add -2 to 34.

$$x=\frac{32}{-6}$$

Reduce the fraction \frac{32}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{16}{3}$$

Now solve the equation x=\frac{-2±34}{-6} when ± is minus. Subtract 34 from -2.

$$x=-\frac{36}{-6}$$

Divide -36 by -6.

$$x=6$$

The equation is now solved.

$$x=-\frac{16}{3} x=6$$

Steps for Completing the Square

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.

$$-3x^{2}+2x-1=-97$$

Add 1 to both sides of the equation.

$$-3x^{2}+2x-1-\left(-1\right)=-97-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$-3x^{2}+2x=-97-\left(-1\right)$$

Subtract -1 from -97.

$$-3x^{2}+2x=-96$$

Divide both sides by -3.

$$\frac{-3x^{2}+2x}{-3}=-\frac{96}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{2}{-3}x=-\frac{96}{-3}$$

Divide 2 by -3.

$$x^{2}-\frac{2}{3}x=-\frac{96}{-3}$$

Divide -96 by -3.

$$x^{2}-\frac{2}{3}x=32$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=32+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=32+\frac{1}{9}$$

Add 32 to \frac{1}{9}.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{289}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{289}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{289}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{17}{3} x-\frac{1}{3}=-\frac{17}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=6 x=-\frac{16}{3}$$