Question

write the discriminant of the quadratic equation (x+5)^2 = 2(5x-3)​

Answer 1

Anwer is in the picture

Hope this helps :)

Answer 2

Answer:

Step-by-step explanation:

Equation

$(x+5)^{2}=2(5x-3)\\\\\Rightarrow{x}^{2}+2\times{x}\times5+5^{2}=10x-6\\\\\Rightarrow{x}^{2}+10x+25-10x+6=0\\\\\Rightarrow{x}^{2}+31=0$

For Quadratic Equation$ax^{2}+bx+c=0 \text{ [where x is the variable and a, b and c are known values]}$

the value of $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$b^{2}-4ac \text{ is called Discriminant (D)}$

when Discriminant (D) is positive, we get two Real solutions  for x

when Discriminant (D) is zero we get just ONE real solution (both answers are the same)

when Discriminant (D) is negative we get a pair of Complex solutions

In above equation  

$D=b^{2}-4ac\\\\\Rightarrow{D}=0-4\times1\times31\\\\\Rightarrow{D}=-124$

As the Discriminant (D) is negative we get a pair of Complex solutions

therefore the value of x is either  

$x=\frac{-b+\sqrt{D}}{2a}$

or

$x=\frac{-b-\sqrt{D}}{2a}$

$x=\frac{-b+\sqrt{D}}{2a}=\frac{+\sqrt{-124}}{2}=\frac{+2\sqrt{31}i}{2}=+\sqrt{31}i$$x=\frac{-b+\sqrt{D}}{2a}=\frac{-\sqrt{-124}}{2}=\frac{-2\sqrt{31}i}{2}=-\sqrt{31}i$