Here is the solution to the quadratic equation: Solve 3x² − 20 = 0 Step 1: Add 20 to both sides of the equation to isolate the term with x². 3x² = 20 Step 2: Divide both sides by 3 to isolate x². x² = 20/3 Step 3: Take the square root of both sides. Remember to include both the positive and negative roots. x = ±√(20/3) Step 4: Rationalize the denominator by multiplying the numerator and denominator by √(3). x = ±(√(20))/(√(3)) × (√(3))/(√(3)); x = ±(√(20 × 3))/(3); x = ±(√(60))/(3); Step 5: Simplify the radical √(60). Since 60 = 4 × 15, we have √(60) = √(4 × 15) = 2√(15). x = ±(2√(15))/(3); The solutions are x = ±(2√(15))/(3).

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3x2 − 20 = 0

July 25, 2026Mathematics

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Answer

x = \pm\frac{2\sqrt{15}}{3}

Step-by-step solution

5 steps

Here is the solution to the quadratic equation:

Solve $3x^2 - 20 = 0$

Add 20 to both sides of the equation to isolate the term with $x^2$ .
$3x^2 = 20$
Divide both sides by 3 to isolate $x^2$ .
$x^2 = \frac{20}{3}$
Take the square root of both sides.
Remember to include both the positive and negative roots.
$x = \pm\sqrt{\frac{20}{3}}$
Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$ .
$x = \pm\frac{\sqrt{20}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$ $x = \pm\frac{\sqrt{20 \times 3}}{3}$ $x = \pm\frac{\sqrt{60}}{3}$
Simplify the radical $\sqrt{60}$ .
Since $60 = 4 \times 15$ , we have $\sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$ .
$x = \pm\frac{2\sqrt{15}}{3}$

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3x2 − 20 = 0

Answer

Step-by-step solution

Add 20 to both sides of the equation to isolate the term with x2x^2x2.

Divide both sides by 3 to isolate x2x^2x2.

Take the square root of both sides.

Rationalize the denominator by multiplying the numerator and denominator by 3\sqrt{3}3​.

Simplify the radical 60\sqrt{60}60​.

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More Mathematics Questions

Add 20 to both sides of the equation to isolate the term with $x^2$ .

Divide both sides by 3 to isolate $x^2$ .

Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$ .

Simplify the radical $\sqrt{60}$ .