# What are the real solution of equation? 2x^2+6x+11=6 ?

neela | High School Teacher | (Level 3) Valedictorian

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To find the real solution of equation? 2x^2+6x+11=6 .

We rewrite the equation by subtracting 6 from both sides.

2x^2+6x+11-6 = 0

2x^2+6x+5 = 0

The discriminant of the equation is (coefficient of x)^2 - 4*(coefficient of x^2)*(constant term) = 6^2-4*2*5 = 36 - 40 = -4 which is negative.

Since the discriminant is negative both roots are not real.

Therefore the the quadratic equation has no real solution as its both roots are not real.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll move all terms to the left side:

2x^2 + 6x + 11 - 6 = 0

We'll combine like terms:

2x^2 + 6x + 5 = 0

Now, we'll verify if the equation has real solutions. For this reason, we'll calculate the discriminant of the equation.

delta = b^2 - 4ac, where ab,c, are the coefficients of the equation:

ax^2 + bx + c = 0

We'll identify a,b,c:

a = 2

b = 6

c = 5

delta = 36 - 40 = -4 < 0

Since delta is negative, then the equation has no real roots, but it has complex roots.