How do you solve the quadratic using the quadratic formula given $x^2-3x+10=0$ ?

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Answer 1 · 2016-11-09T04:37:33

$0=x^2-3x+10$

Quadratic formula:
$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=[-(-3)+-sqrt((-3)^2-4(1)(10))]/[2(1)]$

Simplify:
$x=[3+-sqrt(-31)]/[2]$

$x=[3+-isqrt31]/2$

Answer 2 · 2016-11-09T04:52:40

$x=(3+sqrt(31)i)/2, (3-sqrt(31)i)/2$

$x^2-3x+10=0$ is a quadratic equation in the form $ax^2+bx+c$ , where $a=1$ , $b=-3$ , and $c=10$ .

Quadratic Formula

$x=(-b+sqrt(b^2-4ac))/(2a)$

$x=(-(-3)+-sqrt((-3)^2-4*1*10))/(2*1)$

$x=(3+-sqrt(9-40))/2$

$x=(3+-sqrt(-31))/2$

$x=(3+sqrt(31)i)/2, (3-sqrt(31)i)/2$

How do you solve the quadratic using the quadratic formula given x^2-3x+10=0x^2-3x+10=0?