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Simply by using the formula x= -b +/- Underroot of (b2 - 4.a.c) / 2.a 2+x-6=2x+4

Q: How do you solve the following quadratic equations with the India method A. x2-2x-13 equals 0 B. 4x2-4x plus 3 equals 0 C. x2 plus 12x-64 equals 0 D. 2x2-3x-5 equals 0?

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Two equations: a linear one in y and a quadratic in x.

( x - 2)( x - 6) = 0 x = 2 or x = 6

x^2 -8x-4=0 x=[8±√(64+16)]/2 x=(8±4√5)/2 x=4±2√5

X= (3/5 , -2)

an equation has an equals sign.

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Two equations: a linear one in y and a quadratic in x.

It means you are required to "solve" a quadratic equation by factorising the quadratic equation into two binomial expressions. Solving means to find the value(s) of the variable for which the expression equals zero.

How about finding the solutions of the quadratic equation: x^2-14x+49 = 0

no only equations with x2 and lower powers can be considered quadratic. those with x3 cannot be considered quadratic, just as x2 cannot be considered linear

x = y = 3

( x - 2)( x - 6) = 0 x = 2 or x = 6

If "equations-" is intended to be "equations", the answer is y = -2. If the first equation is meant to start with -3x, the answer is y = 0.2

How many solutions are there to the following system of equations?2x - y = 2-x + 5y = 3if this is your question,there is ONLY 1 way to solve it.

{-1,-2}

X= (3/5 , -2)

x^2 -8x-4=0 x=[8±√(64+16)]/2 x=(8±4√5)/2 x=4±2√5

No. [ y = 4x2 ] is a quadratic equation.