What are the solutions to the equation 4x^2-36x=-93 ?

The solutions to the equation 4x^2-36x=-93 are x= 9/2+sqrt(3)i,x= 9/2-sqrt(3)i

Find the zeros of 4x^2-36x=-93

The zeros of 4x^2-36x=-93 are x= 9/2+sqrt(3)i,x= 9/2-sqrt(3)i

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4 x^{2} - 36 x = - 93

x = \frac{9}{2} + 3 i, x = \frac{9}{2} - 3 i

Solution steps

4 x^{2} - 36 x = - 93

Move

93

to the left side

4 x^{2} - 36 x + 93 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 4 \cdot 93}{2 \cdot 4}

Simplify

(- 36)^{2} - 4 \cdot 4 \cdot 93 : 83 i

x_{1, 2} = \frac{- ( - 36 ) \pm 8 3 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 36 ) + 8 3 i}{2 \cdot 4}, x_{2} = \frac{- ( - 36 ) - 8 3 i}{2 \cdot 4}

x = \frac{- ( - 36 ) + 8 3 i}{2 \cdot 4} : \frac{9}{2} + 3 i

x = \frac{- ( - 36 ) - 8 3 i}{2 \cdot 4} : \frac{9}{2} - 3 i

The solutions to the quadratic equation are:

x = \frac{9}{2} + 3 i, x = \frac{9}{2} - 3 i

lo g_{8} (3 x + 7) < lo g_{8} (7 x + 4)

6 x^{2} - 23 x - 4

\frac{4 ^{5}}{4 ^{2}}

(- 4) y - 3 = 0

2 x y

What are the solutions to the equation 4x^2-36x=-93 ?
The solutions to the equation 4x^2-36x=-93 are x= 9/2+sqrt(3)i,x= 9/2-sqrt(3)i
Find the zeros of 4x^2-36x=-93
The zeros of 4x^2-36x=-93 are x= 9/2+sqrt(3)i,x= 9/2-sqrt(3)i