How do you solve #2x^2 = 14x + 20#?

Question

Eli Assouline · Accepted Answer

#x = (7 +- sqrt(89))/2#

#=> 2x^2 = 14x + 20#

Divide both sides by #2#

#=> x^2 = 7x + 10#

Rearrange the equation

#=> x^2 - 7x - 10 = 0#

It’s in the form of #ax^2 + bx + c = 0#

where,

Use formula for quadratic equation to find #x#

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

#x = (-(-7) +- sqrt((-7)^2 - (4 × 1 × -10)))/(2 × 1)#

#x = (7 +- sqrt(49 + 40))/2#

#x = (7 +- sqrt(89))/2#

Hazel Anglin · Answer

#(7\pm\sqrt(89))/2#

Move everything to left side: #2x^2-14x-20=0#
Factor out a 2: #2(x^2-7x-10)=0#
Solve the parenthetical quadratic (see here for steps)

Isaiah Anthony · Answer

undefined To solve the quadratic equation $2x^2 = 14x + 20$, follow these steps:

1. Rewrite the equation in standard form, setting it equal to zero: $2x^2 - 14x - 20 = 0$.

2. Simplify the equation if necessary.

3. If possible, factor the quadratic expression. If factoring is not feasible, use the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$.

4. Substitute the coefficients $a$, $b$, and $c$ into the quadratic formula and solve for $x$.

5. After obtaining the solutions for $x$, check if they satisfy the original equation by substituting them back into $2x^2 = 14x + 20$.

6. If the solutions are valid, then they are the solutions to the equation. If not, discard them.

By following these steps, you can determine the solutions to the quadratic equation $2x^2 = 14x + 20$.