Question

Hello, I need help with this math problem please

Answer 1

Answer:

4c^2+7c-5=0

4c^2+7c=5

4c^2+7c=5

4c^2+7c-5=0\quad :\quad c=\frac{-7+\sqrt{129}}{8},\:c=\frac{-7-\sqrt{129}}{8}\quad \left(\mathrm{Decimal}:\quad c=0.54472\dots ,\:c=-2.29472\dots \right)

Hope This Helps!!!

Answer 2

Answer:

C) c = -7 ± √129 / 8

Step-by-step explanation:

x = (-b ± √ (b² – 4ac) ) / 2a

[quadratic formula]

where ax² + bx + c = 0.

[quadratic / square trinomial]

given 4c² + 7c – 5 = 0,

↓ ↓ ↓

a b c

where c = x,

c = (-b ± √ (b² – 4ac) ) / 2a.

c = (-(7) ± √ ((7)² – (4)(4)(-5)) ) / 2(4)

c = (-7 ± √ (49 – (-80)) ) / 8

c = (-7 ± √ (129) ) / 8

c = -7 ± √129 / 8