Solve (x)^(2)-5\times+4=0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$${ x }^{ 2 } -5 \times +4=0$$

Solve for x

$x=2\sqrt{5}\approx 4.472135955$
$x=-2\sqrt{5}\approx -4.472135955$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Multiply $5$ and $4$ to get $20$.

$$x^{2}-20=0$$

Add $20$ to both sides. Anything plus zero gives itself.

$$x^{2}=20$$

Take the square root of both sides of the equation.

$$x=2\sqrt{5}$$ $$x=-2\sqrt{5}$$

Steps Using the Quadratic Formula

Multiply $5$ and $4$ to get $20$.

$$x^{2}-20=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $-20$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{0±\sqrt{0^{2}-4\left(-20\right)}}{2}$$

Square $0$.

$$x=\frac{0±\sqrt{-4\left(-20\right)}}{2}$$

Multiply $-4$ times $-20$.

$$x=\frac{0±\sqrt{80}}{2}$$

Take the square root of $80$.

$$x=\frac{0±4\sqrt{5}}{2}$$

Now solve the equation $x=\frac{0±4\sqrt{5}}{2}$ when $±$ is plus.

$$x=2\sqrt{5}$$

Now solve the equation $x=\frac{0±4\sqrt{5}}{2}$ when $±$ is minus.

$$x=-2\sqrt{5}$$

The equation is now solved.

$$x=2\sqrt{5}$$ $$x=-2\sqrt{5}$$