(2x^2+5x-3)=(3x+7)

Solution for (2x^2+5x-3)=(3x+7) equation:

(2x^2+5x-3)=(3x+7)

We move all terms to the left:

(2x^2+5x-3)-((3x+7))=0

We get rid of parentheses

2x^2+5x-((3x+7))-3=0


We calculate terms in parentheses: -((3x+7)), so:

(3x+7)

We get rid of parentheses

3x+7

Back to the equation:

-(3x+7)


We get rid of parentheses

2x^2+5x-3x-7-3=0

We add all the numbers together, and all the variables

2x^2+2x-10=0

a = 2; b = 2; c = -10;
Δ = b2-4ac
Δ = 22-4·2·(-10)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{21}}{2*2}=\frac{-2-2\sqrt{21}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{21}}{2*2}=\frac{-2+2\sqrt{21}}{4}
$