Solve 100-3x+4x-3x^2

$$100-3x+4x-3 { x }^{ 2 }$$

Factor

$$-3\left(x-\frac{1-\sqrt{1201}}{6}\right)\left(x-\frac{\sqrt{1201}+1}{6}\right)$$

Steps Using the Quadratic Formula

Combine -3x and 4x to get x.

$$factor(100+x-3x^{2})$$

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-3x^{2}+x+100=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-3\right)\times 100}}{2\left(-3\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-3\right)\times 100}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-1±\sqrt{1+12\times 100}}{2\left(-3\right)}$$

Multiply 12 times 100.

$$x=\frac{-1±\sqrt{1+1200}}{2\left(-3\right)}$$

Add 1 to 1200.

$$x=\frac{-1±\sqrt{1201}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-1±\sqrt{1201}}{-6}$$

Now solve the equation x=\frac{-1±\sqrt{1201}}{-6} when ± is plus. Add -1 to \sqrt{1201}.

$$x=\frac{\sqrt{1201}-1}{-6}$$

Divide -1+\sqrt{1201} by -6.

$$x=\frac{1-\sqrt{1201}}{6}$$

Now solve the equation x=\frac{-1±\sqrt{1201}}{-6} when ± is minus. Subtract \sqrt{1201} from -1.

$$x=\frac{-\sqrt{1201}-1}{-6}$$

Divide -1-\sqrt{1201} by -6.

$$x=\frac{\sqrt{1201}+1}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1-\sqrt{1201}}{6} for x_{1} and \frac{1+\sqrt{1201}}{6} for x_{2}.

$$-3x^{2}+x+100=-3\left(x-\frac{1-\sqrt{1201}}{6}\right)\left(x-\frac{\sqrt{1201}+1}{6}\right)$$

Evaluate

$$100+x-3x^{2}$$

Solution Steps

Combine -3x and 4x to get x.

$$100+x-3x^{2}$$