What are the solutions to the equation 4x^2-16x-(-20)=0 ?

The solutions to the equation 4x^2-16x-(-20)=0 are x=2+i,x=2-i

Find the zeros of 4x^2-16x-(-20)=0

The zeros of 4x^2-16x-(-20)=0 are x=2+i,x=2-i

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4 x^{2} - 16 x - (- 20) = 0

x = 2 + i, x = 2 - i

Solution steps

4 x^{2} - 16 x - (- 20) = 0

Refine

4 x^{2} - 16 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 4 \cdot 20}{2 \cdot 4}

Simplify

(- 16)^{2} - 4 \cdot 4 \cdot 20 : 8 i

x_{1, 2} = \frac{- ( - 16 ) \pm 8 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 8 i}{2 \cdot 4}, x_{2} = \frac{- ( - 16 ) - 8 i}{2 \cdot 4}

x = \frac{- ( - 16 ) + 8 i}{2 \cdot 4} : 2 + i

x = \frac{- ( - 16 ) - 8 i}{2 \cdot 4} : 2 - i

The solutions to the quadratic equation are:

x = 2 + i, x = 2 - i

x^{2} - 16 x + 61 = 0

(2 x + 1)^{2} = x^{2} + 1

x^{2} - \frac{2}{3} x - \frac{7}{16} = 0

0.51 = \frac{12.96 + 28.09 - c ^{2}}{28.16}

m - 2 = m^{2} + 2 m + 4

What are the solutions to the equation 4x^2-16x-(-20)=0 ?
The solutions to the equation 4x^2-16x-(-20)=0 are x=2+i,x=2-i
Find the zeros of 4x^2-16x-(-20)=0
The zeros of 4x^2-16x-(-20)=0 are x=2+i,x=2-i