Question

quadratic equation. x square +8+20

Answer 1

your equation will be x²+8x-20=0
x²+8x-20=0
x²+10x-2x-20=0
x(x+10)-2(x+10)=0
(x+10)(x-2)=0
x=-10,2

Answer 2

Hey there!

Solving it by applying quadratic formula for equation "x^2 + 8 + 20 = 0"

For a required quadratic equation of $\bf{ax^2 + bx + c = 0}$ the solutions for which can be represented by the quadratic formula of :

$\boxed{\bf{x_{1, \: 2} = \dfrac{- b +- \sqrt{b^2 - 4ac}}{2a}}}$

Here, a = 1,  b = 8,  c = 20.

Solving for positive and negative values respectively:

$\bf{x_{1} = \dfrac{- 8 + \sqrt{8^2 - 4(1)(20)}}{2(1)}}$

By Applying imaginary number rule that is $\bf{\sqrt{- 1} = i}$.

$\bf{x_{1} = \dfrac{- 8 + \sqrt{16} i}{2}}$

$\bf{x_{1} = \dfrac{- 8 + 4 i}{2}}$

$\bf{x_{1} = \dfrac{- 8 + 4 i}{2}}$

$\bf{x_{1} = \dfrac{4(- 2 + i)}{2}}$

$\bf{x_{1} = 2(- 2 + i)}$

$\bf{x_{1} = - 4 + 2 i}$

For the second solution in negative form of equation:

$\bf{x_{2} = \dfrac{- 8 - \sqrt{8^2 - 4(1)(20)}}{2(1)}}$

By Applying imaginary number rule that is $\bf{\sqrt{- 1} = i}$.

$\bf{x_{2} = \dfrac{- 8 - \sqrt{16} i}{2}}$

$\bf{x_{2} = \dfrac{- 8 - 4 i}{2}}$

$\bf{x_{2} = \dfrac{- 8 - 4 i}{2}}$

$\bf{x_{2} = - \dfrac{4(2 + i)}{2}}$

$\bf{x_{2} = - 2(2 + i)}$

$\bf{x_{2} = - 4 - 2 i}$

Therefore the final solutions for this quadratic equations are:

$\boxed{\underline{\bf{x_{1} = - 4 + 2 i}}}$

$\boxed{\underline{\bf{x_{2} = - 4 - 2 i}}}$

Hope this helps you and solves the doubts for getting the equation solved by quadratic method!