Solve 20x^2-60x+45

$$20 { x }^{ 2 } -60x+45$$

Factor

$$5\left(2x-3\right)^{2}$$

Solution Steps

Factor out 5.

$$5\left(4x^{2}-12x+9\right)$$

Consider 4x^{2}-12x+9. Use the perfect square formula, a^{2}-2ab+b^{2}=\left(a-b\right)^{2}, where a=2x and b=3.

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

Rewrite the complete factored expression.

$$5\left(2x-3\right)^{2}$$

Steps Using Square Of Binomial

This trinomial has the form of a trinomial square, perhaps multiplied by a common factor. Trinomial squares can be factored by finding the square roots of the leading and trailing terms.

$$factor(20x^{2}-60x+45)$$

Find the greatest common factor of the coefficients.

$$gcf(20,-60,45)=5$$

Factor out 5.

$$5\left(4x^{2}-12x+9\right)$$

Find the square root of the leading term, 4x^{2}.

$$\sqrt{4x^{2}}=2x$$

Find the square root of the trailing term, 9.

$$\sqrt{9}=3$$

The trinomial square is the square of the binomial that is the sum or difference of the square roots of the leading and trailing terms, with the sign determined by the sign of the middle term of the trinomial square.

$$5\left(2x-3\right)^{2}$$

Steps Using the Quadratic Formula

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.

$$20x^{2}-60x+45=0$$

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.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 20\times 45}}{2\times 20}$$

Square -60.

$$x=\frac{-\left(-60\right)±\sqrt{3600-4\times 20\times 45}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-60\right)±\sqrt{3600-80\times 45}}{2\times 20}$$

Multiply -80 times 45.

$$x=\frac{-\left(-60\right)±\sqrt{3600-3600}}{2\times 20}$$

Add 3600 to -3600.

$$x=\frac{-\left(-60\right)±\sqrt{0}}{2\times 20}$$

Take the square root of 0.

$$x=\frac{-\left(-60\right)±0}{2\times 20}$$

The opposite of -60 is 60.

$$x=\frac{60±0}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{60±0}{40}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and \frac{3}{2} for x_{2}.

$$20x^{2}-60x+45=20\left(x-\frac{3}{2}\right)\left(x-\frac{3}{2}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-60x+45=20\times \frac{2x-3}{2}\left(x-\frac{3}{2}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-60x+45=20\times \frac{2x-3}{2}\times \frac{2x-3}{2}$$

Multiply \frac{2x-3}{2} times \frac{2x-3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-60x+45=20\times \frac{\left(2x-3\right)\left(2x-3\right)}{2\times 2}$$

Multiply 2 times 2.

$$20x^{2}-60x+45=20\times \frac{\left(2x-3\right)\left(2x-3\right)}{4}$$

Cancel out 4, the greatest common factor in 20 and 4.

$$20x^{2}-60x+45=5\left(2x-3\right)\left(2x-3\right)$$

Evaluate

$$5\left(2x-3\right)^{2}$$