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(2x+3)(2x-1)=10
We move all terms to the left:
(2x+3)(2x-1)-(10)=0
We multiply parentheses ..
(+4x^2-2x+6x-3)-10=0
We get rid of parentheses
4x^2-2x+6x-3-10=0
We add all the numbers together, and all the variables
4x^2+4x-13=0
a = 4; b = 4; c = -13;
Δ = b2-4ac
Δ = 42-4·4·(-13)
Δ = 224
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{14}}{2*4}=\frac{-4-4\sqrt{14}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{14}}{2*4}=\frac{-4+4\sqrt{14}}{8} $
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