Solving Quadratics by Factoring and Square Roots

$$x^2+2x=15$$

Subtract 15 before applying the zero product property.

$$x^2+2x-15=0\quad\Longrightarrow\quad(x-3)(x+5)=0$$

The factors have product −15 and sum 2.

$$x=3\quad\text{or}\quad x=-5$$

Each value makes the original left side equal 15.

$$x^2-30=7x\quad\Longrightarrow\quad x^2-7x-30=0$$

Subtract 7x from both sides.

$$(x-10)(x+3)=0\quad\Longrightarrow\quad x=10\;\text{or}\;x=-3$$

Do not set factors of an expression equal to a nonzero right side separately.

$$3x^2-40x=-100\quad\Longrightarrow\quad3x^2-40x+100=0$$

The board corrects the right side to −100. Use that corrected equation.

$$3x^2-30x-10x+100=3x(x-10)-10(x-10)$$

Split −40x into −30x and −10x.

$$(x-10)(3x-10)=0\quad\Longrightarrow\quad x=10\;\text{or}\;x=\frac{10}{3}$$

Solve both linear equations and check both values.

$$x^2-49=0\quad\Longrightarrow\quad(x-7)(x+7)=0$$

Both factors are necessary.

$$x=\pm7$$

The negative value works because (−7) squared is 49.

$$x^2+25=0\quad\Longrightarrow\quad x^2=-25$$

There are no real solutions. This statement concerns real numbers; complex-number solutions are a separate topic.

$$4x^2-5=0\quad\Longrightarrow\quad x^2=\frac54$$

Isolate the square first.

$$x=\pm\frac{\sqrt5}{2}$$

Alternatively factor over the reals as (2x−√5)(2x+√5)=0.

$$x^2=50\quad\Longrightarrow\quad x=\pm5\sqrt2$$

The square root principle works even when the radicand is not a perfect square.