Question

25x²+20x-7=0
solve it.​

Answer 1

Answer:

ajakshdkjeo126889020273

Step-by-step explanation:

zhkakyn

use sjoaooajags the wmkzka w

skaiaik168@

use q18920

18903+178929999+729918929

189939+728*

Answer 2

To find the nature of the roots, find

$b^{2} - 4ac$. If $b^{2} - 4ac$ is less than zero, then it will not have real roots instead it will have imaginary roots or complex roots.

If it is equal to zero, then the roots will be real and equal.

If it is greater than zero, then the roots will be distinct and real.

So, here, $b^{2} - 4ac$ = $20^{2}$ - 4 × 25 × (-7)

$b^{2}$ - 4ac = 400 + 700 > 0

So, the given equation has 2 real and distinct solutions or roots.

The roots are given by the equation,

x = (-b ± $\sqrt{b^{2} - 4ac }$ ) ÷ 2a

where, a is the coefficient of $x^{2}$, b is the coefficient of $x$ and c is the constant.

Here,

x = ( - 20 ± $\sqrt{20^{2} - 4* 25* (-7)}$ ) ÷ 2 * 25

SOLVING,

x = 0.321 or x = -1.121

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