2(x^2+5x)+7=0

Solution for 2(x^2+5x)+7=0 equation:

2(x^2+5x)+7=0

We multiply parentheses

2x^2+10x+7=0

a = 2; b = 10; c = +7;
Δ = b2-4ac
Δ = 102-4·2·7
Δ = 44
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{11}}{2*2}=\frac{-10-2\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{11}}{2*2}=\frac{-10+2\sqrt{11}}{4}
$